Solution:

a. If $x_{n}$ must be constant, then in particular it must hold that
$$x_{1}=x_{0}=a.$$
But then
$$a=x_{1}=2-x_{0}^{2}=2-a^{2},$$
so $a$ must satisfy the equation
$$a^{2}+a-2=0$$
which has as solutions $a=-2$ and $a=1$. No other value can make the sequence constant (but a priori we do not yet know whether these actually work).
Let us verify that both of these values do indeed generate a constant sequence: since the sequence is defined by recursion, at this point we will also have
$$x_{2}=2-x_{1}^{2}=2-a^{2}=a$$
and, in the same way, $x_{3}=a$ and so on (we have in fact given a proof by induction, although one so immediate that it does not need the usual formalization).
Therefore the required values are exactly $-2$ and $1$.

b. Let us prove that this is true for $y=-2$. First of all, observe that since the square of a real number can never be negative, we will have $x_{n} \leq 2$ for every $n \geq 1$; hence, apart from the possible exception of $x_{0}=a$, all values of the sequence are smaller than $2$.
Suppose now that
$$|a|<|y|=|-2|=2,$$
we prove by induction on $n$ that
$$\left|x_{n}\right|<2 \text{ for every } n \geq 0.$$
The base step is exactly the hypothesis $\left(\left|x_{0}\right|=|a|<2\right)$.
Assuming the inequality holds for $x_{n-1}$, we now prove that it also holds for $x_{n}$.
Indeed if $\left|x_{n-1}\right|<2$, evidently $x_{n-1}^{2}<4$; but
$$x_{n}=2-x_{n-1}^{2}$$
and therefore
$$-2=2-4<2-x_{n-1}^{2}=x_{n}<2,$$
that is $\left|x_{n}\right|<2$, which completes the proof of the inductive step.
Incidentally we observe that the property does not hold instead for the other value from point a), $1$: indeed for example if $a=0$ we have $|a|<1$, but $\left|x_{1}\right|=\left|2-a^{2}\right|=2>1$. One can also show that in fact $-2$ is the only real number that has the property indicated in the question.

c. Suppose now that $|a|>2$; it will be more convenient to prove by induction on $n$ the following property, which is stronger than our claim:
$$\text{ "for every } n \geq 0,\left|x_{n}\right|>2 \text{ and } x_{n+1}<x_{n} \text{ ". }$$
To prove the base step we must show that the inequalities hold in the case $n=0$. We observe that the first inequality is exactly our hypothesis, so it certainly holds; for the second, instead, let us set $c_{0}=|a|-2>0$. Then $|a|=2+c_{0}$ and
$$x_{1}=2-a^{2}=2-\left(2+c_{0}\right)^{2}=-2-2 c_{0}-c_{0}^{2}<-(2+c_{0})=-|a| \leq a=x_{0},$$
so the base step is proved.
The proof of the inductive step is entirely analogous. For every $n \geq 1$ suppose that the inequalities are satisfied for $n-1$; then we can set $c_{n-1}=\left|x_{n-1}\right|-2>0$ and derive as above
$$x_{n}=2-x_{n-1}^{2}=-2-2 c_{n-1}-c_{n-1}^{2}<-(2+c_{n-1})=-\left|x_{n-1}\right|$$
(which shows in particular that $x_{n}<0$ as soon as $n \geq 1$).
Then on one hand for all $n \geq 1$
$$\left|x_{n}\right|=-x_{n}>\left|x_{n-1}\right|>2$$
that is the first inequality to be proved;
on the other hand for $n \geq 2$ (that is, when we know that $x_{n-1}<0$)
$$x_{n}< -\left|x_{n-1}\right|=x_{n-1},$$
while for $n=1$ in any case (as we have seen)
$$x_{n}< -\left|x_{n-1}\right| \leq x_{n-1},$$
so the second inequality is also verified for $n$ as well; and by induction at this point the inequalities hold for all $n \geq 0$, that is, the property is true.
But as we said the property is stronger than our claim, which is therefore also proved.