I know that $F\subset F(a)$ and $F\subset F(b)$ are the Galois extensions of degree $n$ and $m$ respectively, $(n,m)=1$.
1) How can I show that $F\subset F(a,b)$ is the Galois extension either?
2) How can I prove that $G(F(a,b)/F)\cong G(F(a)/F) \times G(F(b)/F)$?
So far:
1) I tried to check if $F\subset F(a,b)$ is algebraic, normal and separable.
It's algebraic since finite. It's normal iff it's a splitting field for some polynomial, so I take the polynomial $fg$, where $f$ is the minimal polynomial for $a$ and $g$ is the minimal polynomial for $b$. But why is it separable?
I am not sure if it's useful, but I've proved moreover that $F(a)\cap F(b)=F$ so $a\notin F(b)$ and $b\notin F(a)$.
2) I have no idea how to do this part.
I would be grateful for any help.
Let $f, g$ be the minimal polynomials of $a, b$ over $F$.
Since $F(a)/F$ is Galois, the roots of $f$ are distinct, and similarly those of $g$. You have already noted that $f$ and $g$ have no roots in common. Therefore $F(a, b)$ is the splitting field of the polynomial $fg$, which has distinct roots, and thus $F(a,b)/F$ is a Galois extension, with $\lvert F(a, b) : G \rvert = mn$.
Let
Restrictions give us a group homomorphism $$ \phi : G \to H \times K. $$ Now note that $\lvert G \rvert = m n = \lvert H \rvert \cdot \lvert K \rvert$. And $\phi$ is injective, as if $g \in G$ acts trivially on both $F(a)$ and $F(b)$, it has to be trivial.
Hence $\phi$ is an isomorphism.