Suppose $F: M \to N$ is a local diffeomorphism between smooth manifolds.
$(a)$ Let $G: P \to M$ be a continuous map between smooth manifolds. Then $G$ is smooth if and only if $F \circ G$ is smooth.
$(b)$ Let $H: N \to P$ be any map between smooth manifolds and in addition assume $F$ is surjective. Then $H$ is smooth if and only if $H\circ F$ is smooth.
Is it possible to strengthen this result? Perhaps by dropping the requirement that $G$ is continuous in $(a)$ or $F$ surjective in $(b)$?
You can't drop these two assumptions.
For $(a)$ take $F:\Bbb R\to \Bbb S^1$ defined by $F(t)=e^{2\pi it}$. Then you can check that $F$ is a local diffeomorphism. If you take $G:\Bbb R\to \Bbb R$ defined by $G(t)=\lfloor t\rfloor $ then $G$ is not continuous but $F\circ G(t)=1$ for all $t$ fo $F\circ G$ is smooth.
For $(b)$ take $M=(0,1)$, $N=\Bbb R$ and $F:M\to N$ the inclusion map. Then $F$ is a local diffeomorphism. Now take $H:\Bbb R\to \Bbb R$ which is the identity on $(0,1)$ and which is not nice on the rest of $\Bbb R$, like $H\vert_{\Bbb R-(0,1)}=1_{\Bbb Q}$. Then $H\circ F=F$ is smooth but $H$ is not.
If you prove $(a)$ and $(b)$ you will see why these two hypothesis are necessary.