Question: Find the maximal ideals for $R=$
a) $\Bbb Z/5\Bbb Z$, b) $\Bbb Z/10\Bbb Z$ and c) $\Bbb Z/4\Bbb Z\times\Bbb Z/6\Bbb Z$
My attempt: Since all those rings are finite and commutative, the set of prime ideals is equal to the set of maximal ideals.
For a) the only prime ideal is $(0)$ because $n\Bbb Z/5\Bbb Z$ for $n=2,3,4$ spans $1_R=1+5\Bbb Z=3\cdot(2+5\Bbb Z)=2\cdot(3+2+5\Bbb Z)=4\cdot(4+5\Bbb Z)$
b) $5\Bbb Z/10\Bbb Z$ and $2\Bbb Z/10\Bbb Z=4\Bbb Z/10\Bbb Z=6\Bbb Z/10\Bbb Z=8\Bbb Z/10\Bbb Z$ are the two only maximal ideals.
Here $(0)\subsetneq I\subsetneq R$ is not maximal and also not prime because for example $(2+10\Bbb Z)\cdot(5+10\Bbb Z)=0+10\Bbb Z$
c) $2\Bbb Z/4\Bbb Z\times2\Bbb Z/6\Bbb Z$ and $2\Bbb Z/4\Bbb Z\times3\Bbb Z/6\Bbb Z$ are the only two maximal ideals
$2\Bbb Z/4\Bbb Z\times2\Bbb Z/6\Bbb Z$ and $2\Bbb Z/4\Bbb Z\times3\Bbb Z/6\Bbb Z$ are not maximal ideals of $\Bbb Z/4\Bbb Z\times\Bbb Z/6\Bbb Z$.
Modding out by the first one, you get $\mathbb Z_2\times\mathbb Z_3$, which clearly isn't a field. I'll let you work out the quotient by the second one to see it has the same flaw.
As I'm sure you will have no problem discovering, there are three maximal ideals:
$$2\Bbb Z/4\Bbb Z\times\Bbb Z/6\Bbb Z\\ \Bbb Z/4\Bbb Z\times3\Bbb Z/6\Bbb Z\\\Bbb Z/4\Bbb Z\times2\Bbb Z/6\Bbb Z$$
In general, the maximal ideals of a finite product of commutative rings $\prod_{i=1}^n R_i$ are of the form $\prod_{i=1}^n I_i$ where $I_i\lhd R_i$ and for exactly one $j$, $I_j$ is a prime ideal of $R_j$ and $I_i=R_i$ for all $i\neq j$.