Answer
(a) $$(AB)^{-1}
=\left[ \begin {array}{ccc} -6&-25&24\\ -6&10&7
\\ 17&7&15\end {array} \right]
$$
(b) $$(A^T)^{-1}=\left[ \begin {array}{ccc} 1&0&4\\ -4&1&2
\\ 2&3&1\end {array} \right]
$$
(c) $$(2A)^{-1}
=\left[ \begin {array}{ccc} \frac{1}{2}&-2&1\\ 0&\frac{1}{2}&\frac{3}{2} \\ 2&1&\frac{1}{2}\end {array} \right].$$
Work Step by Step
(a) $$(AB)^{-1}=B^{-1}A^{-1}=\left[ \begin {array}{ccc} 6&5&-3\\ -2&4&-1
\\ 1&3&4\end {array} \right]
\left[ \begin {array}{ccc} 1&-4&2\\ 0&1&3
\\ 4&2&1\end {array} \right]
=\left[ \begin {array}{ccc} -6&-25&24\\ -6&10&7
\\ 17&7&15\end {array} \right]
$$
(b) $$(A^T)^{-1}=(A^{-1})^T=\left[ \begin {array}{ccc} 1&0&4\\ -4&1&2
\\ 2&3&1\end {array} \right]
$$
(c) $$(2A)^{-1}=\frac{1}{2}A^{-1}=\frac{1}{2}\left[ \begin {array}{ccc} 1&-4&2\\ 0&1&3
\\ 4&2&1\end {array} \right]
=\left[ \begin {array}{ccc} \frac{1}{2}&-2&1\\ 0&\frac{1}{2}&\frac{3}{2} \\ 2&1&\frac{1}{2}\end {array} \right].$$