The first statement is a generalized statement of the Inverse Galois problem. For this, the field $F$ is fixed and cannot be chosen. The second statement removes the restriction: $F$ can be any field.
If we set $F = \mathbb{Q}$, the first statement is unknown save for a few choices of $G$. On the other hand, this statement fails when $F$ is a finite field because every Galois extension will have a cyclic Galois group: only $\mathbb{Z}_n$ will appear as a Galois group (for any $n \in \mathbb{N})$.
However, the second statement is true for any finite group $G$. To see this, first recall Cayley's theorem, which states that any finite group $G$ appears as a subgroup of $S_n$ for $n \geq |G|$. Next, Hilbert proved that $S_n$ appears as a Galois group over $\mathbb{Q}$; let $K$ be a Galois extension with $\text{Gal}(K/\mathbb{Q}) \cong S_n$. Finally, the Galois correspondence tells us that, for each subgroup $H$ of $S_n$, there will exist an intermediate field $\mathbb{Q} \subset E_H \subset K$ such that the subgroup is isomorphic to $\text{Gal}(K/E_H)$. This is to say, $G$ will appear as the Galois group for the extension $K/E_G$.