= Solution
The polynomial $f(x)=x^5+2x+2$ satisfies the <Eisenstein criterion> at the prime two: every nonleading coefficient is divisible by two, and the constant coefficient is not divisible by four. Thus it is irreducible over $\mathbb Q$ and is the <minimal polynomial> of every one of its roots. Part (a) gives
$$
\mathbb Z[\alpha]/(5)\cong\mathbb F_5[x]/(\overline f).
$$
But
$$
f(x)=(x-1)(x^4+x^3+x^2+x+3)+5.
$$
In the quotient, the two nonzero classes of $x-1$ and $x^4+x^3+x^2+x+3$ multiply to zero. They are nonzero because their nonzero representative degrees are less than five. The quotient is not an <integral domain>, so \b[$(5)$ is not a <prime ideal>].
For <ring homomorphisms> preserving the identity, an image $\beta$ of $\alpha$ in the <Gaussian integers> would satisfy $f(\beta)=0$. The <minimal polynomial> of $\beta$ over $\mathbb Q$ would then be $f$, of degree five. This is impossible because $\beta\in\mathbb Q(i)$ has degree at most two. Thus \b[there is no unital homomorphism $\mathbb Z[\alpha]\to\mathbb Z[i]$]. If homomorphisms are instead allowed not to preserve the identity, the zero map exists; every nonzero homomorphism into this integral domain is unital, because the image of one is a nonzero idempotent and hence one.
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