a.
Note that the numbers are
$$\begin{aligned} a_k + (-1)^{k+1}(k+1) &= \binom{p-2}{k} + (-1)^{k+1}(k+1) \\ &= \frac{(p-2)(p-3)\cdots(p-1-k)}{k!} + (-1)^{k+1}(k+1) \end{aligned}$$
where $1 \le k \le p-2$. Since
$$(p-2)(p-3)\cdots(p-1-k) \equiv (-2)(-3)\cdots(-1-k) = (-1)^k(k+1)! \pmod{p}$$
and $p \nmid k!$, we have
$$a_k + (-1)^{k+1}(k+1) \equiv (-1)^k(k+1) + (-1)^{k+1}(k+1) = 0 \pmod{p}.$$

b.
The answer is 1728.
For each choice of $x_1, x_2, x_3$ satisfying the inequalities, we have
$$x_4 = 100 - x_1 - x_2 - x_3 \ge 100 - 10 - 11 - 12 > 50.$$
Therefore, it suffices to count the number of triples $(x_1, x_2, x_3)$ satisfying the inequalities. The answer is
$$(10 - (-1) + 1)(11 - 0 + 1)(12 - 1 + 1) = 12^3 = 1728.$$