Yes. Let $M$ be the cyclic maximal subgroup (actually teh the proof work for an Abelian maximal subgroup). We may suppose by induction that $M$ contains no non-trivial normal subgroup of $G.$ Then for each non-identity subgroup $X$ of $M,$ we have $M = N_{G}(X),$ as $M$ is maximal and $X \lhd M.$ It follows easily that $M$ is a Hall subgroup of $G,$ and by Burnside's normal $p$-complement theorem, there is a normal complement $K$ to $M$ in $G.$ Furthermore, we have $M = C_{G}(m)$ for each non-identity $m \in M.$ Hen $G$ is a Frobenius group with Frobeius kernel $K$ and Frobenius complement $M.$ Since Frobenius kernels are nilpotent by Thompson's theorem, we see that $G$ is solvable as $K$ and $G/K \cong M$ both are. In fact, I think this argument (for the case of $M$ Abelian) is due to Thompson.
Yes. Let $M$ be the cyclic maximal subgroup (actually teh proof work for an Abelian maximal subgroup). We may suppose by induction that $M$ contains no non-trivial normal subgroup of $G.$ Then for each non-identity subgroup $X$ of $M,$ we have $M = N_{G}(X),$ as $M$ is maximal and $X \lhd M.$ It follows easily that $M$ is a Hall subgroup of $G,$ and by Burnside's normal $p$-complement theorem, there is a normal complement $K$ to $M$ in $G.$ Furthermore, we have $M = C_{G}(m)$ for each non-identity $m \in M.$ Hen $G$ is a Frobenius group with Frobeius kernel $K$ and Frobenius complement $M.$ Since Frobenius kernels are nilpotent by Thompson's theorem, we see that $G$ is solvable as $K$ and $G/K \cong M$ both are. In fact, I think this argument (for the case of $M$ Abelian) is due to Thompson.