First one seems right.
Second one:
How can you make a 0/1 matrix of order 2 negative ? By ensuring the determinant evaluates as $0 - 1 = -1$
i.e. The leading diagonal has to multiply to 0, and other diagonal has to multiple to 1.
Therefore, the leading diagonal can be (0,0), (0,1),(1,0) and the other diagonal is (1,1).
So I think out of 16 possible combinations, 3 combinations will yield a negative determinant:
$\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$, $\begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix}$, $\begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix}$.
Probability of a negative determinant $= \frac{3}{16}$
Probability of a positive determinant $= 1- \frac{3}{16} = \frac{13}{16}$
What is the right answer ?
Edit: my definition of positive is 'non-negative', so I've included $0$ too