Regarding 3), this "Big Picard" stuff is serious overkill.
Think like an undergraduate real analysis student:
The p-series zeta(p) converges for real p > 1, whereas zeta(1) = sum of the harmonic series = oo.
An easy argument using (e.g.) the integral test shows that
lim_{p -> oo} zeta(p) = 1.
The function zeta(p) is continuous in p [the convergence is uniform on right half-planes, hence on compact subsets], so by the intermediate value theorem it takes on every positive integer value n >= 2 at least once -- and, since it is a decreasing function of p, exactly once -- on the real line.
Thus A_n is nonempty for all n >= 2.
EDIT: Let me show that zeta(s) takes on all real values infinitely many times on the negative real axis.
For this, note that for all n >=1,
zeta(-(2n-1)) = - B_{2n}/(2n),
where B_{2n} is the (2n)th Bernoulli number. It is known that the B_{2n}'s alternate in sign and grow rapidly in absolute value:
|B_{2n}| \sim 4 \sqrt{\pi n} (n/(pi e))^{2n}.
The claim follows from this and the Intermediate Value Theorem.
