I think you are asking whether the Mori cone NE(X) on a surface X is closed. A ruled surface counterexample due to Mumford is given in Lazarsfeld's Positivity in Algebraic Geometry I, Example 1.5.1.

If $X$ is a projective toric variety then $A_0(X) = \mathbb{Z}$ (this is true because it's resolution is rationally connected), and then the localization exact sequence implies that for any nontrivial open subset $A_0(U) = 0$.

I think the extension sequence you start with only applies when $X$ is factorial. Otherwise the Jacobian is an abelian variety already, not a semi-abelian variety.

I believe it is not known whether exceptional divisors for resolutions of rational singularities are always acyclic, as this can depend on singularities of the exceptional divisors themselves. What is known is that for a snc resolution, the fibers are acyclic, see Lemma 2.5 in arxiv.org/pdf/2212.06786.pdf.

See (4.1) and the following Lemma in Karmazyn-Kuznetsov-Shinder, "Derived categories of singular surfaces".

If the fiber X has simple normal crossings, then cohomology $H^{>0}(X_{red}, \mathcal{O}) = 0$, for the reduced fiber, and this can be proved using Hodge theory see e.g. arxiv.org/pdf/2212.06786.pdf, Lemma 2.5.

As in Sasha's comment, toric degenerations are too restrictive for these questions (in type II for K3s elliptic curves appear as intersections); usually one considers snc or toroidal. In arxiv.org/pdf/1503.08320.pdf, Question 7, it is asked if the dual complex is always homeomorphic to a sphere, as soon as it has dimension n. See (34) in that paper where it's explained why it's at least homology sphere.

A very recent excellent overview "On rationality problems" by Debarre: perso.imj-prg.fr/olivier-debarre/wp-content/uploads/sites/34‌​/… No smooth rational hypersurfaces of degree 4 and higher are known, and this is one of the key questions in the field, alongside rationality of cubic fourfold. Welcome to MathOverflow!

It seems a lot easier to use the whole derived pushforward (as a complex, not just the cohomology sheaves), so that if $\pi$ is a resolution and $\sigma$ is a smooth blow up, then $R(\pi\sigma)_*(\mathcal{O}) = R\pi_* R\sigma_*(\mathcal{O}) = R\pi_*(\mathcal{O})$. Using Weak factorization (in char. 0) this implies that the complex $R\pi_*(\mathcal{O})$ on a given singular variety $X$ is independent of the choice of the resolution $\pi$.

Tony was claiming the other direction, I think, if points are separated in some weak sense, the variety is quasi-affine, which seems nontrivial.

Regarding point separation, if $U \subset X$ is open, and regular functions separate points on $X$, then surely they separate points on $U$, just because restriction of a regular function is regular?

I suppose by applying the automorphis $x \mapsto x + c$ we can assume that $p_1 = p_2 = 0$. Then it should be the same result as over algebraically closed field with the same proof.