I was unsure if I should post this because the question has not been active for a year. I decided to go ahead anyway, because people might have the same question in the future.

The following paragraph is taken from Structure of Lie Groups and Lie Algebras by
V. V. Gorbatsevich, A. L. Onishchik and E. B. Vinberg.

Let $A$ be a simply-connected covering group for $SL_2(\mathbb{R})$. As is
known, $Z(A) \simeq \mathbb{Z}$. Let $z_0$ be a generator of the group $Z(A)$, and let $t_0 \in \mathbb{T}=SO_2$ be a rotation through an angle incommensurable with $\pi$. Then $\Gamma = \langle(t_0, z_0)\rangle$ is a discrete normal subgroupof $\mathbb{T} \times A$. Let $G = (\mathbb{T} \times A)/\Gamma$. Then the image $L$ of the subgroup A under the natural homomorphism $\mathbb{T} \times A \rightarrow G$ is a Levi subgroup of G. The fact
that the subgroup $\langle t_0 \rangle$ is dense in $\mathbb{T}$ implies that $L$ is dense in $G$, i.e. it is not a Lie subgroup.

This shows that question (1), which is true for algebraic groups in characteristic 0 as explained in the other answer, is not generally true for Lie groups, as the Levi factor can fail to be closed.