Solution:

(i) (AB)$^{t}$ = B$^{t}$ A$^{t}$

Taking Left Hand Side: (AB)$^{t}$

= ${{\left( \left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right] \right)}^{t}}$

= $\left[ \begin{matrix} -10 & 2 \\ -17 & 4 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: B$^{t}$ A$^{t}$

= ${{\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]}^{t}}\times {{\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 1 & -3 \\ 2 & -5 \\ \end{matrix} \right]\times \left[ \begin{matrix} -1 & 2 \\ 3 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1\times \left( -1 \right)+\left( -3 \right)\times 3 & 1\times 2+\left( -3 \right)\times 0 \\ 2\times \left( -1 \right)+\left( -5 \right)\times 3 & 2\times 2+\left( -5 \right)\times 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & 2+0 \\ -2-15 & 4+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & 2 \\ -17 & 4 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that: (AB)$^{t}$ = B$^{t}$ A$^{t}$

(ii) (BC)$^{t}$ = C$^{t}$B$^{t}$

Taking Left Hand Side: (BC)$^{t}$

= ${{\left( \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]~\times ~\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]~ \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} 1\times \left( -2 \right)+2\times 3 & 1\times 6+2\times \left( -9 \right) \\ -3\times \left( -2 \right)+\left( -5 \right)\times 3 & -3\times 6+\left( -5 \right)\times \left( -9 \right) \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} -2+6 & 6-18 \\ 6-15 & -18+45 \\ \end{matrix} \right] \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} 4 & -12 \\ -9 & 27 \\ \end{matrix} \right] \right)}^{t}}$

= $\left[ \begin{matrix} 4 & -9 \\ -12 & 27 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: C$^{t}$B$^{t}$

= ${{\left[ \begin{matrix} -2 & 6 \\ 3 & -9 \\ \end{matrix} \right]}^{t}}\times {{\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} -2 & 3 \\ 6 & -9 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & -3 \\ 2 & -5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2\times 1+3\times 2 & -2\times \left( -3 \right)+3\times \left( -5 \right) \\ 6\times 1+\left( -9 \right)\times 2 & 6\times \left( -3 \right)+\left( -9 \right)\times \left( -5 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2+6 & 6-15 \\ 6-18 & -18+45 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & -9 \\ -12 & 27 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that: (BC)$^{t}$ = C$^{t}$B$^{t}$