Suppose $A\subset B$ is an inclusion of commutative rings with $B$ Jacobson.
If $B$ is finitely generated as an algebra over $A$ does it follow that $A$ is Jacobson?
If $B$ is finitely generated as a module over $A$ does it follow that $A$ is Jacobson?
For the modulefinite (i.e. second) question:
Yes, it is true: The point is that $A \subseteq B$ is integral, and the claim is true for all integral extensions.
When $A \subseteq B$ is integral, every prime $\mathfrak{p} \subseteq A$ comes as $\mathfrak{P}\cap A$ for a prime $\mathfrak{P} \subseteq B$ ("lying over theorem"), and morever, if $\mathfrak{P}$ is maximal then so is $\mathfrak{p}$ (e.g. by "going up theorem").
So given a prime $\mathfrak{p}$ of $A$, choose such lift $\mathfrak{P} \subseteq B$. By $B$ being Jacobson one has $\mathfrak{P}=\bigcap_{\mathfrak{P}\subseteq{\mathfrak{M}}\subseteq_{\mathrm{max}}B}\mathfrak{M}$. Intersecting this with $A$ yields $$\mathfrak{p}=\bigcap_{\mathfrak{P}\subseteq{\mathfrak{M}}\subseteq_{\mathrm{max}}B}(\mathfrak{M}\cap A)$$ where all the ideals $\mathfrak{M}\cap A$ are maximal in $A$. Thus, $A$ is Jacobson.

1$\begingroup$ Dear Pavel, maybe it is worth emphasizing from the beginning that your elegant argument proves the result for an arbitrary integral overring $B$ of $A$, not only for finite ones. At my first superficial reading of your post I thought that integrality was only used as a step in the proof, whereas it suffices for the whole proof to go through. Anyway, bravo! (And +1, of course.) $\endgroup$ – Georges Elencwajg Feb 23 at 9:01

$\begingroup$ @GeorgesElencwajg Dear Georges, thank you, I have edited the answer as you proposed. $\endgroup$ – Pavel Čoupek Feb 23 at 18:07

$\begingroup$ Well done Pavel: I hope your answer will get the many more upvotes it deserves $\endgroup$ – Georges Elencwajg Feb 23 at 18:25
A counterexample for the first question is any DVR $R$. Clearly, $R$ is not Jacobson. But if $\pi$ is the uniformizer, then $Q(R) = R[\frac{1}{\pi}]$ is a finitely generated $R$algebra and a field, hence Jacobson.

1$\begingroup$ I would not be surprised if the second statement is true, though. $\endgroup$ – Martin Brandenburg Feb 22 at 21:03