Solution:

a.
If $(x, y)$ is a pair of integers that satisfies the inequality, then $(-x, -y)$ is also such a pair, since
$$(-x)^2 + (-x)(-y) + (-y)^2 = x^2 + x y + y^2.$$
So we can match up pairs of solutions to the inequality, $(x, y) \leftrightarrow (-x, -y)$. Every solution will be paired with a different solution, except for the one remaining solution $(0, 0)$ which is paired with itself. This shows that the number of solutions is odd.

b.
This is similar to the previous part, except now that we have to arrange the nonzero solutions into triples instead of pairs. If $(x, y)$ is a solution to the inequality, then so is $(-x-y, x)$, since
$$(-x-y)^2 + (-x-y)x + x^2 = x^2 + x y + y^2$$
Applying this transformation three times in succession gives the cycle
$$(x, y) \rightarrow (-x-y, x) \rightarrow (y, -x-y) \rightarrow (x, y)$$
so we can unambiguously arrange the solutions into cycles of three, of the form $\{(x, y), (-x-y, x), (y, -x-y)\}$. Now, if any two solutions in the same cycle are equal, then the third is also equal to them, so every cycle contains either three distinct solutions or just one solution. If a cycle contains just one solution $(x, y)$, then $x = y$ and $y = -x - y$ gives $x = y = 0$. Therefore, the solution $(0, 0)$ forms a cycle by itself, and every other cycle consists of three different solutions, which means that the total number of solutions has remainder $1$ when divided by $3$.