This is true if and only if $N_L(G)/G$ has trivial center, where $N_L(G)$ is the normalizer in $L$ of $G$.

EDIT: Realized I had misread the question.

The set of $l$ such that $[l,g]\in G$ for all $G$ is called the normalizer of $G$. It's not particularly common for the normalizer to be the same as G, though it does happen sometimes. There are a few theorems I know about specific Lie algebras being self-normalizing, such as Cartan and Borel subalgebras in reductive Lie algebras, but it's not very common and I don't know any general condition which guarantees it.

This is true if and only if $L/G$ has trivial center.

This is true if and only if $N_L(G)/G$ has trivial center, where $N_L(G)$ is the normalizer in $L$ of $G$.