I'll just address your first question.
Let
$$
h(t) = \begin{matrix} 1 &\; \text{if} \; t \in [0,1] \\ 0 &\; \text{else}\end{matrix}
$$
and let
$$
g_n(t) = (h*h*\cdots*h)(t)
$$
be the n-fold convolution of $h$. Observe that
$$
\widehat{g_n}(w) = \left(\frac{1}{iw}\right)^n\left(1-e^{-iw}\right)^n =: \hat{b}(w)\hat{c}(w)
$$
where $\hat{f}(w) = \int_{\mathbb{R}} f(t) e^{-iwt} dt$. So then by the convolution theorem, we have
$$
g_n(t) = (b*c)(t)
$$
To find $b(t)$ and $c(t)$, you will have to find these distributional inverse transforms, and use the binomial theorem for $\hat{c}(w)$. I'll leave this to you. You should find that:
$$
g_n(t) = \frac{1}{2(n-1)!}\sum_{j=0}^n \begin{pmatrix} n \\ j \end{pmatrix} (-1)^j (t-j)^{n-1} \text{sign}(t-j)
$$
You of course would have to show this is equivalent to your linked question.