Let $i_A : A \to G$ and $i_B : B \to G$ be the embeddings and let $\mu :  G \times G \to G$ be the group multiplication.  Then the composition
$$
A \times B \stackrel{i_A\times i_B}{\longrightarrow} G \times G \stackrel{\mu}{\longrightarrow} G
$$
is a smooth map which is surjective and sends the identity to the identity.  (It is not, however, a group homomorphism.)

In a previous, incorrect, version of this answer, I had claimed that the tangent map at the identity was surjective.  This is of course false.  This was used only to arrive at an inequality:
$$
\dim\mathrm{Lie}(G) \leq \dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B)~.
$$

In fact, this inequality follows from [Sard's theorem][1] and the fact that the map $A \times B \to G$ is surjective, for if $\dim G > \dim A + \dim B$ then the image of the map $A \times B \to G$ would have measure zero.

Having established the above inequality, we now establish the reverse inequality.  To see this simply notice that if $\mathrm{Lie}(A) \cap \mathrm{Lie}(B) = 0$, then the fact that $\mathrm{Lie}(A)$ and $\mathrm{Lie}(B)$ are subspaces of $\mathrm{Lie}(G)$ implies
$$
\dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B) \leq \dim\mathrm{Lie}(G)~.
$$
So the answers to Q1 and Q2 are both true (as vector spaces not as Lie algebras).


  [1]: https://en.wikipedia.org/wiki/Sard%27s_theorem