Dear Nicojo, since you now have many counter-examples, let me give you a situation where $B$ *is* finitely generated, in line with your question 2). I am going to adopt your notations with the important caveat that $k$ is a ring which needn't be a field .

**Theorem of Artin-Tate** Consider the inclusions of rings  $k \subset B  \subset A$ .  Suppose that $k$ is Noetherian, that $A$ is a finitely generated *algebra* over $k$ and that $A$ is a finitely generated *module* over B.Then $B$ is a finitely generated *algebra* over $k$.

You might interpret this as saying that when $B$ is  sufficiently close to $A$, finite generation is preserved.

You can find the proof in Atiyah-Macdonald, Proposition 7.8, page 81.
From this theorem you can then prove Zariski's result that an extension of fields that is finitely generated as an algebra is actually a finite-dimensional extension (Proposition 7.9 page 82 *loc.cit.*) and then  Hilbert's Nullstellensatz is literally an exercise: exercise 14, page 85 . So this result of Artin-Tate is really basic in commutative algebra and algebraic geometry, not surprisingly if you consider the authors (the Artin here is Emil, Mike's father.)