Prove Quotient Ring is Isomorphic to Product of Fields 
Problem: (a) Show $\mathbb{Z}[i]/\langle 5\rangle$ is a product of two fields. (b) Show $\mathbb{Z}[i]/\langle 3\rangle$ is a field. (c) Show $\mathbb{Z}[i]/\langle 2\rangle$ is neither a field nor a product of two fields.

My Attempt: Now from first glance, it is clear (a) and (c) will not be fields since both $5$ and $2$ are not irreducible in $\mathbb{Z}[i]$ and thus will not general maximal ideals and hence the quotient ring cannot be a field. It can also be shown that $\langle 3\rangle$ forms a maximal ideal, thus (b) is a field.
What I am having difficultly with is showing (a) is a product of 2 fields and that (c) is not a product of 2 fields.
For (a), I initially thought of the homomorphism, $\phi: \mathbb{Z}[i] \rightarrow \mathbb{Z}_5 \times \mathbb{Z}_5$ given by $a+bi \rightarrow (a \mod (5), b \mod (5))$ and then thought to show $\ker (\phi ) = \langle 5\rangle$ to prove $\mathbb{Z}[i]/\langle 5\rangle \cong \mathbb{Z}_5 \times \mathbb{Z}_5$. However I soon realised that $\phi$ is not a ring homomorphism since: $$\phi(a+bi)\star \phi(c+di) \neq \phi((a+bi)\cdot (c+di))$$
Question: Could anyone please point me in the right direction for parts (a) and (b).
Also more generally, is there any statement that can be made regarding if $\mathbb{Z}[i]/\langle a\rangle$ (where $a$ is an integer) can be written as some product of field?
Thank you.

Hint:
a) In $\mathbf Z[i]$, one has $5=(2+i)(2-i)$ and these factors are irreducible since their norm is equal to $5$.  Furthermore, $\mathbf Z[i]$ is a P.I.D., so each generates a maximal ideal, and you can apply the Chinese remainder theorem:
$$\mathbf Z[i]/5\mathbf Z[i]\simeq \mathbf Z[i]/(2+i)\times\mathbf Z[i]/(2-i).$$
b) $$\mathbf Z[i]/2\mathbf Z[i]\simeq\mathbf Z[X]/(X^2+1)\!\Bigm/\!\!2\cdot\mathbf Z[X]/(X^2+1)\\\simeq \mathbf Z/2\mathbf Z[X]/(X^2+1)\simeq \mathbf Z/2\mathbf Z[X]/\bigl((X+1)^2\bigr).$$
