Ideals rarely contains $1$ but $1$ always lie in the invariant ring so I don't understand the question.

@ulrich: I think you get an effective bound by using the effective bounds on the order of automorphism groups of curves (by Stichtenoth). As the statement is true if $p$ doesn't divide the order of any automorphism group of a curve (of genus $g$). (On the other hand I think this is if and only if so in order to have it true for large $p$ on needs to know that the automorphisms groups of curves in char. $p$ is not divisible by $p$ for large $p$.)

I think you are misusing the notion of corepresentable (which as I understand it is just the dual of "representable" and hence doesn't even make sense for contravariant functors). So the question should be if $M_g\times\mathbb Z/p$ is the coarse moduli space of the algebraic stack $\mathcal M_g\times\mathbb Z/p$.

Dan is completely right, the only thing you have to do (as Dan indicates) in positive characteristic is to note that first order deformations deformations of principally polarised abelian varieties is a subspace of the first order deformation space of abelian varieties. Hence we get a subspace of $\mathrm{Hom}(H^0(C,\Omega^1),H^1(C,\mathcal O_C))=H^1(C,\mathcal O_C)^{\oplus2}$ consisting of the symmetric tensors. This is $\Gamma^2H^1(C,\mathcal O_C)$ rather than $S^2H^1(C,\mathcal O_C)$.

All of these hyperplanes pass through the origin so as soon as there are an infinite number of roots it is not locally finite.

It is true if $L$ is finitely presented as $R$-module (I don't think finitely generated is enough) and $S$ is flat as $R$-module.

I on the other hand feel that I must disagree with Donu. Applying Hilbert's syszygy theorem to a (finitely generated) graded module whose $\widetilde{(-)}$ is the given $\mathbb F$ gives exactly what the OP asked (by applying $\widetilde{(-)}$ to the resolution). Auslander-Buchsbaum-Serre is generalising Hilbert's theorem to more general situations.

I've added a discussion of formal $\mathbb G_m$-actions.

Sauf erreur the inclusion $BO(k) \to BO$ induces on cohomology the map that sets $w_i$ to $0$ for $i>k$. This map commutes with the action of the Steenrod algebra so you already have a formula.

Note however that diagonalisation is not possible over $\mathbb Z$ (or even $\mathbb Q$) which make things trickier. However, my guess is that the easiest way is to first do it over a splitting field and then descend (at least to $\mathbb Q$, over $\mathbb Z$ there should be even more complications).