Solution:

a) Yes. Here is an example:
$$\left(\begin{array}{rrrr} 1 & 0 & 1 & 1 \\ 1 & -1 & -1 & -1 \\ 1 & -1 & 1 & 0 \\ 1 & -1 & 1 & -1 \end{array}\right)$$

b) No. We have 11 possibilities for these sums: $0, \pm 1, \pm 2, \pm 3, \pm 4$ and $\pm 5$. Denote by $a_{i}$ the sum of the numbers in the $i$-th row and by $b_{j}$ the sum of the numbers in the $j$-th column. Obviously
$$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=b_{1}+b_{2}+b_{3}+b_{4}+b_{5}$$
which shows that the number of odd sums $a_{i}$ and the number of odd sums $b_{j}$ are of the same parity. Therefore all odd sums must be achieved.

Without loss of generality we may assume that $b_{1}=5$ and then none of $a_{i}$ equals $-5$. Thus, we may suppose that $b_{2}=-5$. At least one of the sums equals $4$ or $-4$ and suppose it is $4$. This is possible only if there is a column with four $1$'s and one $0$. Let $b_{3}=4$ and the zero is in the last row. Therefore $a_{i} \neq -3$ for every $i$ and we may assume that $b_{4}=-3$. Then in the fourth column we have at least three $-1$'s. If they are in the first four rows we may assume that they are in the first three rows. So $a_{1}, a_{2}, a_{3}$ is a permutation of $-1, 0, 1$. Therefore $b_{5} \neq 3$ and since $a_{5} \neq 3$, it follows that $a_{4}=3$, i.e., there are $1$'s in the last two cells of the fourth row. Since $b_{4}=-3$, the number in the last cell of the fourth column is $-1$. Now every possibility for the number in the 5th row and 5th column leads to a contradiction.

It remains to consider the case when there are at most two $-1$'s in the first four cells of the fourth column. We may assume that the numbers in the fourth column are consequently $-1, -1, 0, 0$ and $-1$. Since the sum of the first four numbers of the rows equal $0, 0, 1, 1$ and $-1$ we have that none of the rows equals $3$ and therefore $b_{5}=3$. We must use different numbers for rows 1 and 2 and for rows 3 and 4, so we may have at most three $1$'s. Hence we may assume that the numbers in the fifth column are $1, 0, 1, 0, 1$ and then $a_{1}=a_{4}=1$, a contradiction.