Answer
Part A:
$\begin{bmatrix}
7 & 7 & 14\\
8 & 8 & 16\\
-1 & -1 & -2\\
\end{bmatrix}$
Part B:
$\begin{bmatrix}
13\\
\end{bmatrix}$
Part C:
Not possible.
Work Step by Step
A has 3 rows and 1 column (3 x 1).
B has 1 row and 3 columns (1 x 3).
Part A:
$\begin{bmatrix}
7(1) & 7(1) & 7(2)\\
8(1) & 8(1) & 8(2)\\
(-1)(1) & (-1)(1) & (-1)(2)\\
\end{bmatrix}$ = $\begin{bmatrix}
7 & 7 & 14\\
8 & 8 & 16\\
-1 & -1 & -2\\
\end{bmatrix}$
Part B:
$\begin{bmatrix}
1(7) + 1(8) + 2(-1)\\
\end{bmatrix}$ = $\begin{bmatrix}
7 + 8 - 2\\
\end{bmatrix}$ = $\begin{bmatrix}
13\\
\end{bmatrix}$
Part C:
Since the number of columns in A does not equal the number of rows in A, AA is not possible.