$q(x)\equiv x ~(\text{mod}(x^2+2))$
$q(x) \equiv 1 (\text{mod} ~x)$
$q(x) \equiv (x+1) (\text{mod} (x^2 + 2x + 2))$
I know that this is a question about the Chinese remainder theorem but I have no idea how to do it with polynomials!
Should I first add up all the mods ?
I suppose that you could use Chinese Remainder, but since the moduli are not exactly independent, I would just look at it with a piercing eye and see what’s what.
The second requirement, $q\equiv1\pmod x$, simply says that your polynomial has constant term $1$, or in other words that $q(0)=1$. The first requirement requires that $q\equiv x\pmod{x^2-1}$. (Remember that you’re in characteristic three, where $2=-1$.) Since any multiple of $x^2-1$ vanishes at $\pm1$, the condition $q\equiv x\pmod{x^2-1}$ says that $q$ has the same value at $1$ and $-1$ as $x$ does. So the two first conditions specify $q(0)=1$, $q(1)=1$, and $q(-1)=-1$. Surely no degree-one polynomial can satisfy this, and you know that three values determine a parabola. I’ll leave that to you.