I think $S$ has no embedding as a quartic in $\mathbb{P}^3$: we are looking for an integral vector $v = zh - xe_1 - ye_2$ such that $v$ intersects $e_1$, $e_2$, $h - e_1 - e_2$ positively and $v^2 = 4$. From the positivity of the intersection we get $x, y > 0$, $z > x+y$, so that $z^2 - x^2 - y^2 > 2xy$ and from the square $4$ condition $2xy \le z^2 - x^2 - y^2 = 2$, so $xy < 1$ and we don't get any solutions.

Then after changing the basis, the intersection form seems to be diagonal $2(x^2 - y^2 - z^2)$, and one needs to solve $x^2 - y^2 - z^2 = 2$ to find elements of square $4$. One obvious element of square $4$ is $2h - E_1 - E_2$ (pullback from the del Pezzo surface multiplies the degree by two), however it contracts the other $(-2)$-curve (pullback of $h - E_1 - E_2$).

$S$ has three $(-2)$-curves obtained as preimages of lines on the del Pezzo surface. It seems that these curves form a basis of $NS(S)$?

@JasonStarr: I am not sure what you mean exactly; however I don't think we can blow up $\mathbb{A}^n$ to make the quotient by the given linear action smooth. I think it's false already for cyclic quotients of $\mathbb{A}^2$. Probably you meant something else.

@JasonStarr: Larsen-Lunts is for smooth projective varieties, and it's not immediately applicable here, as the quotient is not smooth. That is, if $[U] = \mathbf{L}^n$ for a quasi-projective variety $U$ we may not immediately deduce that $U$ is stably rational.

As an abstract group $Pic^0(C)$ is the same for every genus $g$ curve, because all complex tori of the same dimension are diffeomorphic? On the other hand, genus is recovered, as dimension of the $p$-torsion of this group (over $\mathbb{F}_p$) equals $2g$.

This holds in Bittner's blow up Grothendieck ring (isomorphic to the usual one in char. 0), as blow up relations preserve connected components. In char. p in general one could try to use etale cohomology with compact supports as this is what zeta-functions boil down to over finite fields.

Ah, I see. Maybe one can still hope for commuting non-symplectic involutions? Another idea is to look at multiplication by (-1) for two different elliptic structures on X.

You seem to be essentially asking about two involutions $\sigma_1$, $\sigma_2$ in $O(NS(X))$ which both act by $-1$ on the discriminant lattice, and whose $+$-eigenspaces on $NS(X)$ are distinct?

Seems that property (a) always holds for the following reason: taking the quotient of $X$ by the subgroup of $H$ of $Aut(X)$ generated by $\sigma_1$ and $\sigma_2$, the resulting (singular) surface $X/H$ will have an ample divisor, whose pullback to X will still be ample?

In the case when $X$ is singular, we do not expect to move divisors off singular points, e.g. in a quadric cone $xy = z^2$ lines making up the cone all pass through the singular point $0$ and there is no way to move them off it -- in the sense of linear equivalence, and probably in the sense you are asking as well.

In the parameter space of all lines passing through $p$ (this is a projective space), interesecting a codimension two subvariety away from $p$ is a proper closed condition. Hence we are asking whether this projective space can be covered by countably many proper closed subvarieties, which is not the case. Thus there exsits a line with the required properties.

By projection formula this vanishing holds if $\pi_*(E^i) = 0$ in Chow groups; this must be the case when $i$ is less than codimension of the center of the blowup. On the other hand, when $n = 3$, $i = 2$ and we blow up a curve, the answer is not necessary zero (it is roughly degree of the curve).

A good reference is Orlov's paper on smooth and proper triangulated dg-categories: arxiv.org/abs/1402.7364

Regarding exceptional collections, it not true that if $R$ is a finite dimensional $k$-algebra of finite global dimension (this corresponds to $P$ being smooth proper) has a full exceptional collection in $D^b(R-mod)$.

I think what you ask for is existing of a tilting generator $T$ in $D^b(P)$; then $R = End(T)$ is the corresponding ring, in fact, finite-dimensional noncommutative $k$-algebra.

Welcome to Mathoverflow! This holds for finite fields. For an infinite field a necessary condition could be that every element has roots of arbitrary degree, otherwise you can construct a nontrivial torsor. See here for a discussion and references: mathoverflow.net/questions/377081/… and this paper for a sharp recent result: degruyter.com/document/doi/10.1515/crelle-2016-0037/html