For a metric-compatible connection, and a local orthonormal frame, the matrix representation of the curvature is skew-hermitian. So, if they say hermitian, it's just a typo.
Effects of levels of compatibility and structure on curvature.
Here are some answers I wrote about curvature and bundle-valued forms, which I think you may find helpful:
So, for any smooth vector bundle $(E,\pi,M)$ with a connection $\nabla$, its curvature is a smooth vector bundle morphism $R:\bigwedge^2(TM)\to \text{End}(E)$; in words it is a "$2$-form on $M$ with values in $\text{End}(E)$" or an "$\text{End}(E)$-valued $2$-form on $M$". In other notation, we also write $R\in \Omega^2(M;\text{End}(E))$.
Depending on the level of structure we impose on the vector bundle, and the amount of compatibility we require on the connection, the more we are able to say about its curvature. For example,
If $(E,\pi, M,g)$ is a (real) vector bundle with metric and $\nabla$ is metric-compatible, then the curvature is skew-adjoint-valued, i.e $R:\bigwedge^2(TM)\to \text{Skew}_g(E)$, where for each $x\in M$ $(\text{Skew}_g(E))_x:=\text{Skew}_{g_x}(E_x)$ consists of skew-adjoint endomorphisms relative to $g$, instead of just $R:\bigwedge^2(TM)\to \text{End}(E)$.
Analogously, if $(E,\pi,M,h)$ is a smooth complex vector bundle (not necessarily holomorphic) with hermititan metric $h$ and $\nabla$ is a metric-compatible, then its curvature is skew-hermitian-valued, i.e $R:\bigwedge^2(TM)\to \text{Skew}_h(E)$.
Statement (1) can be proved directly, but see the section "Generalities" below for a more general statement and proof. Statement (2) is a mild variation of (1) so I won't prove it.
If $(E,\pi,M)$ is any holomorphic vector bundle over a complex manifold, with a connection $\nabla$ that is compatible with the holomorphic structure, then the $(0,2)$ grading of the curvature vanishes, i.e restricting the full map
\begin{align}
R:\bigwedge^2(TM)=\bigwedge^{(2,0)}(TM)\oplus \bigwedge^{(1,1)}(TM)\oplus \bigwedge^{(0,2)}(TM)\to \text{End}(E)
\end{align}
to a map $\bigwedge^{(0,2)}\to \text{End}(E)$ yields $0$. The proof of this is pretty clear from Cartan's structure equation for curvature: we have
\begin{align}
\Theta=d\theta+\theta\wedge\theta=\overline{\partial}\theta+\partial\theta+\theta\wedge\theta.
\end{align}
If we started with a local frame of holomorphic sections, then by compatibility of $\nabla$ with the holomorphic structure, it follows all the connection 1-forms $\theta^a_{\,b}$ are usual forms on (an open subset of) $M$ of type $(1,0)$. So, $\Theta$ is a $(1,1)$ form plus a $(2,0)$ form plus a $(2,0)$ form. In particular, there is no $(0,2)$ term. Since this is a well-defined notion, the proof is complete. Notice crucially how compatibility of $\nabla$ with the holomorphic structure comes into play: if not, the $\theta$'s would also have a $(0,1)$ term, which when acted on by $\overline{\partial}$ would produce a $(0,2)$ term.
By putting together points (2) and (3), we see that if $(E,\pi,M,h)$ is a holomorphic vector bundle with hermitian metric, and $\nabla$ is compatible with $h$ and the complex structure, then the curvature is skew-hermitian-valued, and the $(0,2)$ part of it vanishes. But in fact, we can say something more: taking the adjoint relative to $h$ (i.e at the level of matrix representations relative to orthonormal frames, we're taking the conjugate transpose) we see that $(2,0)$ and $(0,2)$ types get flipped. So, in fact, we can deduce that the $(2,0)$ must also vanish. Therefore, the curvature is a form of type $(1,1)$. Putting both these facts together, the curvature in this case must be of type $(1,1)$ and skew-hermitian-valued. In symbols,
\begin{align}
R\equiv \Theta\in \Omega^{(1,1)}\left(M; \text{Skew}_h(E)\right).
\end{align}
(The notation $R$ is classic, and pays homage to Riemann, while $\Theta$ is just a capitalized $\theta$ (it's also common to write $\omega$ for the connection 1-forms and $\Omega$ for the curvature 2-forms) and if you look in gauge theories, you'll see it denoted as $F$... so, don't mind the different letters for the curvature).
Hopefully this answers your main question 5 and hopefully my first link clarifies what bundle-valued forms are.
What kind of object is the curvature?
Next, for your question 4 (mainly 4.1), which is about the 'type' of thins, maybe my answer to Why do we write a vector-valued form as a tensor product? helps. But also, when you go from the level of sections, to the level of bundle-morphisms, we're invoking the 'tensor characterization lemma' (see Lee's books, or look around on the site, because it's asked repeatedly)
So, the following objects are all the 'same' conceptually: for any two vector bundles $E,F$ over $M$,
- a $C^{\infty}(M)$-linear map $\Gamma(E)\to \Gamma(F)$
- a vector bundle morphism $E\to F$
- a smooth section of $\text{Hom}(E,F)$
- a smooth section of $E^*\otimes F$.
So now let $F=\bigwedge^2(T^*M)\otimes E$. Then, the following things are 'equal' (there are many other things which are possible, but I'll just write out some of them):
- a $C^{\infty}(M)$-linear map $\Gamma(E)\to \Gamma(\bigwedge^2(T^*M)\otimes E)$
- a smooth section of
\begin{align}
E^*\otimes \left(\bigwedge^2(T^*M)\otimes E\right)
&\cong \left(\bigwedge^2(T^*M)\right)\otimes (E^*\otimes E)\\
&\cong \left(\bigwedge^2(T^*M)\right)\otimes \text{End}(E)\\
&\cong \left(\bigwedge^2(TM)\right)^*\otimes \text{End}(E)\\
&\cong \text{Hom}\left(\bigwedge^2(TM); \text{End}(E)\right).
\end{align}
- a vector bundle morphism $\bigwedge^2(TM)\to \text{End}(E)$.
Any one of these types of things is called an $\text{End}(E)$-valued $2$-form on $M$, and the space of all such guys is denoted $\Omega^2(M;\text{End}(E))$. This and my links about bundle-valued forms should answer your question 4 fully.
Generalities:
The statement (1) is very classial and is one of the well-known symmetries of the Riemann tensor in differential geometry ($R_{abcd}=-R_{bacd}$). This is actually a consequence of something much more general about the simple nature of the second exterior covariant derivatives. Note, you do NOT need to prove this general result to get what you want; there are several direct arguments, but I figured I'd show this as well to highlight the conceptual simplicity of things:
Lemma. (Easy Leibniz for $d_{\nabla}^2$)
Let $E,F,G$ be smooth vector bundles over the same base manifold $M$, and let $\beta:E\oplus F\to G$ be a smooth bilinear bundle morphism (i.e a 'product' in the sense of my first link). Suppose we have connections $\nabla^E,\nabla^F,\nabla^G$ which are all $\beta$-compatible, in the sense that for all sections $\psi$ of $E$ and $\phi$ of $F$, and all vector fields $X$ on $M$ we have that
\begin{align}
\nabla^G_X\bigg(\beta(\phi,\psi)\bigg)&=\beta(\nabla^E_X\psi,\phi)+ \beta(\psi,\nabla^F_X\phi).
\end{align}
Then,
The second exterior-covariant derivatives satisfy the following nice equation: for all $\psi\in\Omega^k(M;E)$ and all $\psi\in \Omega^l(M;F)$, we have that
\begin{align}
d_{\nabla}^2(\psi\wedge\phi)&=(d_{\nabla}^2\psi)\wedge\phi+\psi\wedge(d_{\nabla}^2\phi).
\end{align}
More precisely, (see my first link for explanation of the notation) is
\begin{align}
d_{\nabla^G}^2(\psi\wedge_{\beta}\phi)&=
(d_{\nabla^E}^2\psi)\wedge_{\beta}\phi
+\psi\wedge_{\beta}(d_{\nabla^F}^2\phi)
\end{align}
Keeping in mind the definition of curvature (see second link) we can write the conclusion of this lemma as
\begin{align}
R_G(\psi\wedge\phi)&=(R_E\wedge\psi)\wedge\phi+ \psi\wedge (R_F\wedge\phi),
\end{align}
or more precisely,
\begin{align}
R_G\wedge_{\text{ev}}(\psi\wedge_{\beta}\phi)&= (R_E\wedge_{\text{ev}}\psi)\wedge_{\beta}\phi+
\psi\wedge_{\beta}(R_F\wedge_{\text{ev}}\phi).
\end{align}
Corollary.
In particular, taking $E=F$ above and $G=M\times \Bbb{R}$ the trivial bundle and $\beta=g$ to be a bundle metric on $E$, and $\nabla$ a metric-compatible connection, and $k=l=0$, we see that the conclusion of (2) above says that for all sections $\psi,\phi$ of $E$,
\begin{align}
0&=g(R\cdot \psi,\phi)+g(\psi,R\cdot \phi),
\end{align}
i.e $R$ is skew-adjoint-valued.
We only need to prove statement (1) in the lemma since (2) follows immediately as I explained; finally the corollary also follows immediately. So, to prove (1), we have
\begin{align}
d_{\nabla}^2(\psi\wedge\phi)&=d_{\nabla}(d_{\nabla}\psi\wedge\phi+ (-1)^k\psi\wedge d_{\nabla}\phi)\\
&=\bigg(d_{\nabla}^2\psi\wedge \phi + (-1)^{k+1}d_{\nabla}\psi\wedge d_{\nabla}\phi\bigg)+ (-1)^k\bigg(d_{\nabla}\psi\wedge d_{\nabla}\phi+ \psi\wedge d_{\nabla}^2\phi\bigg)\\
&=(d_{\nabla}^2\psi)\wedge\phi+ \psi\wedge (d_{\nabla}^2\phi),
\end{align}
where the middle two terms clearly cancel out. This completes the proof of (1). All that's left is verifying that each step I wrote down is actually valid, and understanding rigorously what the $\nabla$ at each stage means (I explained all of this in my very first link about bundle-valued forms).
