Let $\mathfrak{g}$ be a Lie algebra over a field of characteristic $0$. The interesting case for these questions is when $\mathfrak{g}$ is infinite dimensional.
Is $U(\mathfrak{g})$ a coherent algebra?
Are finitely presented $U(\mathfrak{g})$-modules perfect?
Here are some definitions from the questions: an algebra $A$ is (left) coherent if the kernel of any left $A$-module homomorphism $A^{\oplus n} \to A$ is finitely generated. An $A$-module is perfect if it has a bounded resolution by finitely generated projective $A$-modules.
Note that the second question is stronger than the first: a positive answer to the second question would imply a positive answer to the first one.
Some motivation: by the PBW theorem, enveloping algebras can be thought of as analogous to polynomial algebras. Polynomial algebras, even infinitely generated ones, satisfy both of the above conclusions.
The answers are positive in the following cases:
$\mathfrak{g}$ is finite-dimensional. Then $U(\mathfrak{g})$ is Noetherian with finite global dimension.
$\mathfrak{g}$ is abelian. Then $U(\mathfrak{g})$ is a polynomial algebra.
$\mathfrak{g}$ is free. Then $U(\mathfrak{g})$ is a tensor algebra, so has global dimension 1, which can be seen to give positive answers to both questions.
For what it's worth, I'm not optimistic about either question, but would be glad to at least learn counterexamples.
