Answer
See below
Work Step by Step
a) $BC=\begin{bmatrix}
0 & ab\\ 0 & 0
\end{bmatrix} \\
CB=\begin{bmatrix}
0 & ab\\ 0 & 0
\end{bmatrix}$
Hence, $BC=AB$
b) $C^2=\begin{bmatrix}
0+0 & 0+0\\ 0+0 & 0+0
\end{bmatrix}=0_2\\
e^{Ct}=I_2+Ct+\frac{1}{2!}(Ct)^2\\
=I_2+Ct+0+0...\\=\begin{bmatrix}
1 & 0\\ 0 & 1
\end{bmatrix}+\begin{bmatrix}
0 & bt\\ 0 & 0
\end{bmatrix}\\
=\begin{bmatrix}
1 & bt\\ 0 & 1
\end{bmatrix}
$
c) Since $e^{A+B}t=e^{At}e^{Bt}\\
\rightarrow e^{At}=e^{(B+C)t}\\
=e^{Bt}e^{Ct}\\
=\begin{bmatrix}
e^{at} & 0\\ 0 & e^{at}
\end{bmatrix}\begin{bmatrix}
1 & bt\\ 0 & 1
\end{bmatrix}\\
=\begin{bmatrix}
e^{at} & e^{at}bt\\ 0 & e^{at}
\end{bmatrix}$
