Solution:

a) If $a=0$, then $x=-y$ and hence $2x^{2}=2$. It follows that $(x, y) = (1, -1)$ or $(x, y) = (-1, 1)$.

b) We know from a) that $a=0$ is one of the desired numbers. Let $a \neq 0$. Setting $x+y=p$, $xy=q$, we have $p=a q$ and $p^{2}-2q=a^{2}+2$. Then $a^{2} q^{2}-2q-a^{2}-2=0$ and hence $(p, q)=(-a, -1)$ or $(p, q)=\left(\dfrac{a^{2}+2}{a}, \dfrac{a^{2}+2}{a^{2}}\right)$. Note that these pairs of numbers are different. The first case $x+y=-a$, $xy=-1$ leads to the quadratic equation $z^{2}+a z-1=0$ which has two distinct real roots $z_{1}$ and $z_{2}$. It follows that $(x, y) = (z_{1}, z_{2})$ and $(x, y) = (z_{2}, z_{1})$ are solutions of the given system. Thus we have to find all $a$ for which the second case is impossible. This means that the discriminant of the quadratic equation $z^{2}-\dfrac{a^{2}+2}{a} z+\dfrac{a^{2}+2}{a^{2}}=0$ is negative, i.e., $a \in (-\sqrt{2}, \sqrt{2}) \setminus \{0\}$. So the answer to b) is $a \in (-\sqrt{2}, \sqrt{2})$.