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(1) Suppose a (finite-dimensional, real) Lie group $G$ is connected and simply-connected. Can $G$ have a non-trivial central extension by a discrete group, i.e. can there exist a Lie group $\tilde G$ such that $G \simeq \tilde G/Z$ where $Z$ is a discrete central subgroup of $\tilde G$, and where $\tilde G$ is not just a direct product $G\times Z$?

(2) Related question: Suppose $G$ is a Lie group that is connected, simply-connected and semisimple. Can it have any non-trivial central extensions at all?

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    (1) no: indeed then $\tilde{G}=Z\times\tilde{G}_0$ and $\tilde{G}_0\to G$ is a topological isomorphism. (2) is a bit imprecise. If you also mean in the category of Lie groups the answer is no too (use Levi factors). – YCor Feb 24 '20 at 07:20

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(1) Such an extension can not exist if you want $\widetilde{G}$ to be connected, because then $G=\widetilde{G}/Z$ for discrete $Z$ implies $Z\subset\pi_1G$.

To see this, take $z\in Z$, a basepoint $\tilde{x}_0\in\widetilde{G}$ and a curve connecting $\tilde{x}_0$ to $z\cdot\tilde{x}_0$. Its image is a loop in $G$, which is not 0-homotopic. Indeed, if it was, then by the homotopy lifting property, the curve in $\widetilde{G}$ would be homotopic to a point by a homotopy which fixes $Z\cdot \tilde{x}_0$ as a set. Because this set is discrete, it would have to be fixed pointwise. So you get a 0-homotopy of a curve that fixes both endpoints, which is absurd.

If $\widetilde{G}\not=G\times Z$, then at least some connected component of $\widetilde{G}$ is preserved by some nontrivial subgroup of $Z$ and you can apply the above argument to that component.

(2) If $Z$ is not discrete, then Lie group extensions correspond to Lie algebra extensions, which are classified by $H^2(\mathfrak{g},\mathfrak{z})$. The latter is zero for semisimple Lie algebras, by the second Whitehead Lemma.

user39082
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    (1) $\pi_0$ applied to $1\to Z\to\tilde G\to G\to1$ gives an isomorphism $Z\to\pi_0\tilde G$, whence a morphism $\tilde G\to Z$ which splits the sequence. –  Feb 24 '20 at 07:20