I deleted my previous comment about intersection points in an affine chart; what it said was correct but misleading. I now claim you need $q$ to be at least 19 to get all intersections in one affine chart. This follows essentially from the Hasse bound. I don't know if this is sharp.

If this map is not dominant, then $X$ must be a cone with vertex at $x$.

The question is settled in the negative but I just mention a maybe relevant source of counter-examples, namely the so-called parabolic automorphisms of algebraic surfaces. The following paper of J. Grivaux gives a useful exposition: arxiv.org/pdf/1307.1771

To (over-)elaborate a bit on your second paragraph: a picture I find useful is to view $H^2(X,\mathbb Z)$ as a fixed lattice inside $H^2(X,\mathbb C)$, while $H^{1,1}(X)$ is a subspace that is "moving around" inside $H^2(X,\mathbb C)$. As we deform $X$ and this subspace "moves around" it sometimes "lands on" more of the lattice points, meaning that the rank of NS jumps up.

Your argument in the last paragraph seems to assume that any dense subset of $S$ contains a nonempty open set. That is not true.

I forgot to say something about the second question. The answer to that is "no"; if $Y$ is an elliptic curve and $X=Y \times Y$ then its Picard group has rank at least 3 (cf. Hartshorne Ex IV.4.10.)

@KConrad: the mistaken comment is now deleted. Thank you for this correction.

@TabesBridges: I think $n-O(1)$ means a function of $n$ which differs from $n$ by a constant not depending on $n$ (nothing to do with the line bundle $O(1)$).

One software that can do this is Magma: magma.maths.usyd.edu.au/magma/handbook/text/1804 This is not freely available, but it has an online calculator here: magma.maths.usyd.edu.au/calc Beware though that this is limited to "small" calculations. A freely available offline alternative is Normaliz: normaliz.uni-osnabrueck.de

Yes, if you know the generators of a rational polyhedral cone, you (or a software) can figure out the generators of its dual. (This much has nothing to do with algebraic geometry.) Is that the kind of answer you are looking for?

It's not true that $-K_X$ spans an extremal ray of the effective cone: it is an ample line bundle, so it must be in the interior of that cone (even the nef cone). If you view $X$ as a blowup of $\mathbf P^2$ then I think you can get "new" divisors in $|-2K_X|$ for example as sums $C_i+E_i$ where $E_i$ is an exceptional divisor over one of the points $p_i$, and $C_i$ is a sextic which is double at all the points and triple at $p_i$. I did not translate this to your coordinates but maybe you can do this.

What about a cuspidal curve?

This is because for example there is an isomorphism $X \cong \mathbf P^1 \times \mathbf P^1$ which takes the ruling to one of the $\mathbf P^1$-fibres. It is clear that one can move such a fibre to a curve that is disjoint, hence $E^2=0$.

No. $E$ is a line in one of the two rulings of $X$ so it satisfies $E^2=0$, but if $X \rightarrow Y$ is a contraction to another surface then all contracted curves must have negative self-intersection.

In my copy (2nd ed.) Chapter 8 is titled "Moduli" and Chapter 9 is the one you mention.

Birkenhake--Lange has a lot of material on this topic, in the complex setting.