An eigenvector is a vector such that $T \vec v=\lambda \vec v$. This means that such a vector, when transformed by $T$, change its length but non chang its direction, i.e. $T \vec v$ stay on the same straight line as $\vec v$.
Helping with a drawing you can easily see that the eigenvectors of your transformation can only stay on the given line $r$ of equation $y=mx$ or on its orthogonal line passing by the origin, i.e. the line $y=-\frac{1}{m}x$. All other vectors in the plane, when reflected on $r$, ''jump'' on an other straight line. And since
$$
\vec v_1=
\left[
\begin {array}{ccccc}
1\\
m\\
\end {array}
\right]
\qquad
\vec v_2=
\left[
\begin {array}{ccccc}
-m\\
1\\
\end {array}
\right]
$$
are vectors on those two lines, they are eigenvectors.
Now note that $\vec v_1$ is a fixed point of $T$ so that its eigenvalue must be $\lambda_1=1$ and $\vec v_2$ is changed by $t$ in a vector that has opposite coordinates, so that its eigenvalue is $ \lambda_2=-1$.
The matrices of your second question is:
$T=PAP^{-1}$ where $A$ is the diagonal matrix having $a_{11}=\lambda_1$ and $a_{22}=\lambda_2$, and $P$ is the matrix that have as columns the eigenvectors, so you have:
$$
P=
\left[
\begin {array}{ccccc}
1&-m \\
m& 1
\end {array}
\right]
\qquad
P^{-1}=
\dfrac{1}{1+m^2}
\left[
\begin {array}{ccccc}
1&m \\
-m& 1
\end {array}
\right]
$$
$$
A=
\left[
\begin {array}{ccccc}
1&0 \\
0& -1
\end {array}
\right]
$$
and if you perform the product $T=PAP^{-1}$ you find the searched matrix.
To complete the exercise use $m=\tan \theta$ to show that the matrix $T$ has the form:
$$
T=
\left[
\begin {array}{ccccc}
\cos 2\theta&\sin 2\theta \\
\sin 2\theta& -\cos 2\theta
\end {array}
\right]
$$
that is the general form of a reflection in the plane.
