Let $G_1$ and $G_2$ be finite solvable groups and $M$ be a maximal subgroup of $G=G_1\times G_2$ show that one of the following holds: $$ G_1\times\{e\}\leq M,\ \{e\}\times G_2\leq M,\ M \unlhd G.$$
We know that $G$ is solvable then since $M\leq G$, $M$ is also solvable. And $[G:M]=p^k$ for some prime p.(Because maximal subgroup of a solvable group has prime power index.)
I couldn't make them useful for this question.