Let $m,n\in \mathbb N$ and $A,B\in M_n(\mathbb R)$ so that: $$A^mBA^n=I$$ $$B=\begin{bmatrix} \;1&-1&\;3&\;1\\\;1&\;1&\;2&\;1\\\;2&-1&\;3&\;2\\-1&-2&\;1&\;2\end{bmatrix}$$ $(a)$ Is $A$ regular?
$(b)$ Calculate $A^{m+n}$
$(a)$ $A$ is regular because $\det {(A^mBA^n)}=\det A\cdot\det {(A^{m-1}BA^n)}\ne0$
$(b)$ $A^{m+n}=A^mIA^n=A^m(B\frac{1}{B})A^n=\frac{1}{B}$
I inverted $B$ blockwise.
As on Wikipedia (in notation here I used $B$ for a block-matrix, not to mix itwith the initial matrix): $${\begin{bmatrix} A&B\\C&D\end{bmatrix}}^{-1}=\begin{bmatrix} A^{-1}+A^{-1}B{(D-CA^{-1}B)}^{-1}CA^{-1} &- A^{-1}B{(D-CA^{-1}B)}^{-1}\\{(D-CA^{-1}B)}^{-1}CA^{-1} & {(D-CA^{-1}B)}^{-1}\end{bmatrix}$$ For the sake of simplicity, I transformed B into upper-triangular matrix: $$B=\begin{bmatrix} \;1&-1&\;3&\;1\\\;1&\;1&\;2&\;1\\\;2&-1&\;3&\;2\\-1&-2&\;1&\;2\end{bmatrix}\rightarrow\begin{bmatrix} \;1&-1&\;3&\;1\\\;0&\;2&-1&\;0\\0&\;1&-3&\;0\\\;0&-3&\;4&\;3\end{bmatrix}\rightarrow\begin{bmatrix} \;1&-1&\;3&\;1\\\;0&\;0&\;5&\;0\\\;0&\;1&-3&\;0\\\;0&\;0&-5&\;3\end{bmatrix}\rightarrow\begin{bmatrix} \;1&-1&\;3&\;1\\\;0&\;1&-3&\;0\\\;0&\;0&5&\;0\\\;0&\;0&-5&\;3\end{bmatrix}\rightarrow\begin{bmatrix} \;1&-1&\;3&\;1\\\;0&\;1&-3&\;0\\\;0&\;0&\;5&\;0\\\;0&\;0&\;0&\;3\end{bmatrix}\rightarrow\begin{bmatrix} \;1&\;0&\;0&\;1\\\;0&\;1&-3&\;0\\\;0&\;0&\;5&\;0\\\;0&\;0&\;0&\;3\end{bmatrix}$$ /edited: here I could've get $I_4$, but I would like to go through different ways/ $$A=I, B=\begin{bmatrix}\;0&\;1\\-3&\;0\end{bmatrix}, C=0_2,D=\begin{bmatrix}\;5&\;0\\\;0&\;3\end{bmatrix}$$
Then, the Schur complement of A: $$D-CA^{-1}B=D$$ I got $$D^{-1}=\begin{bmatrix}\frac{1}{5}&\;0\\\;0&\frac{1}{3}\end{bmatrix}$$ Then$$-BD^{-1}=\begin{bmatrix}\;0&-\frac{1}{3}\\\frac{3}{5}&\;0\end{bmatrix}$$
Finally: $$B^{-1}=\begin{bmatrix}\;1&\;0&\;0&\frac{1}{3}\\\;0&\;1&-\frac{3}{5}&0\\\;0&\;0&\frac{1}{5}&\;0\\\;0&\;0&\;0&\frac{1}{3}\end{bmatrix}$$ This appears to be just a little different than $B$ transformed. Is this correct?
You are correct that $A^{m+n}$ is $B^{-1}$, but your reasoning is incomplete and your numerical answer for $B^{-1}$ is wrong.
You claimed that $A^m(BB^{-1})A^n=B^{-1}$, but why is it true? Since $B^{-1}A^n$ is not necessarily equal to $A^nB^{-1}$, you need to justify the previous equality. One may argue as follows: as $(A^mB)A^n=I$, we have $A^n(A^mB)=I$ as well (because $XY=I$ implies that $YX=I$ for square matrices). Hence $A^{m+n}B=I$, i.e. $A^{m+n}=B^{-1}$.
And your result for $B^{-1}$ is a triangular matrix. This is definitely wrong. In general, for a nonsingular matrix $B$, $B^{-1}$ is triangular if and only if $B$ is triangular, but the $B$ in question is clearly not triangular.