Ideal of $\mathbb{Z}[x]$ generated by $3$ and $x^2+1$ is proper

I'm honestly not sure what you're trying to say in (1).

You're correct in that you want to show that $1\not\in I$, but I'm not sure what the whole thing with $1=h(x)\sum_k 3^{i_k}(x^2+1)^{j_k}$ is, or where it came from. The correct thing to say is that the elements of $I$ are of the form $3f+(x^2+1)g$, with $f,g\in\Bbb{Z}[x]$. Thus if $1\in I$, we have $1=3f+(x^2+1)g$. Reducing this mod 3, we see that $(x^2+1)g \equiv 1 \pmod{3}$, which is nonsense, since $\Bbb{Z}_3$ is a domain (indeed a field), so positive degree polynomials are not units in $\Bbb{Z}_3[x]$. Thus this is impossible.

You have the right idea for 2, but it can be simplified a bit. Note that $f_a(I)$ always contains 3, so to show that $f_a(I)=\Bbb{Z}$, it suffices to show that for any $a$, $f_a(I)$ contains an integer that is $\pm 1$ mod $3$. This is easier than trying to solve for $1$ (though I think that could also be done).

We have two cases:

• $a\equiv 0\pmod{3}$, then $a^2+1\equiv 1\pmod{3}$.
• $a\equiv \pm 1 \pmod{3}$, then $a^2+1\equiv 2\pmod{3}$.

We're done.