3, and thus 2 and 1: yes. By checking equality between the two sets at each point, we reduce to the case where the base is a point. But points are always Noetherian schemes, and the statement is obviously true for Noetherian schemes.

3, and thus 2 and 1: yes. By checking equality between the two sets at each point, we reduce to the case where the base is a point. But points are always Noetherian schemes, and the statement is obviously true for Noetherian schemes.

Edit: I was just reminded of this question and I realize that I now know the answer. If an element is in the kernel of the morphism $A \to \mathcal O_Z(Z)$, then it is in the kernel of the map to $\mathcal O_Z$ restricted to each of the $n+1$ standard affine open sets in $\mathbb P^n_A$. In each of those sets, it is in the ideal generated by the $f_i$. Of course if you are in the ideal generated by the $f_i$ then you are in the ideal generated by finitely many of them. Taking a finite union over the $n+1$ different affine opens, we get a finite set of relations that proves this element is in the kernel. So finite elimination theory generates infinite elimination theory.