No. Try $(x_1-1, x_1x_2)$.

This is certainly false for families of abelian varieties: subvarieties of abelian varieties cannot be deformed. I am not sure about the affine case.

If it's geometrically irreducible and has arithmetic genus 1, it is either smooth, or it is a geometrically rational curve with a single node. This last case is not very hard to analyze.

The reason why Grothendieck works with quotients is that this gives a representable functor for any module, not just finitely generated projective modules.

Is the equality $f^{-1}(p) = \mathbb{P}^1$ true scheme-theoretically?

If you add rational tails you can make the automorphism as large as you want.

The Jacobian of a curve of compact type is the product of the Jacobians of the components, and the polarization is the product of the polarizations. This makes ampleness clear, I think.

Yes, I meant the unique maximal torus in $G$.

Any map from a torus to an abelian variety is trivial (for example, because an abelian variety can not contain any rational curves). This implies that $T$ is the unique maximal torus in $A$, and proves the statement.

I doubt that this question has a sensible answer.

Ben McKay's argument only works in characteristic 0. But it is always true that if you have a non-constant map $X \to Y$ of smooth projective curves, the genus of $X$ is at least equal to the genus of $Y$; see Example 2.5.4. in Hartshorne.

The local rings that you describe are all regular, so any arithmetic surface that is not regular can not have this form. In general you get complete locar rings of the form $R[[S,T]]/(ST-\pi^n)$ for some $n \ge 0$.

I believe that the ring $\mathbb C[x,y](y^2 - x^2(x-1))$ (the ring of the standard affine nodal cubic) should have this property.

One can also prove this with descent theory, without reducing to the normal case. And there is a formal version, that works for any noetherian domain.