Let $f,g \in \mathbb{Q}[x]$ be polynomials such that $\{f(a) : a \in \mathbb{Q}\} \subseteq \{g(a) : a \in \mathbb{Q} \}$. Must there be some $h \in \mathbb{Q}[x]$ such that $f(x) = g(h(x))$ for all $x \in \mathbb{Q}$ ?
We don't need Faltings theorem here. See Theorem 1 in: H. Davenport, D. J. Lewis, A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964), 107116.

2
Yes. Consider the curve $X: f(x)=g(y)$. It has infinitely many rational points so has a component of genus zero or one, by Faltings. But because it has so many points of any given height (any rational is the $x$ coordinate of a point), it has to have a component of genus zero, so there are rational functions $u(x),v(x) \in \mathbb{Q}(x)$ with $f(u)=g(v)$. But your assumption actually implies that $u$ is surjective as a function $\mathbb{Q} \to \mathbb{Q}$. That forces $u$ to be a fractional linear transformation and, inverting it, gives what you want.
Maybe I should worry about the irreducibility of $X$, but I'll let you do that instead.