Given a prime $p$ how many primes $\ell<p$ of a given quadratic character mod $p$?

GH from MO and Anonymous have commented above on (modest) lower bounds for the first problem. Let me mention here that a version of problem 2 (of producing many non-residues) appeared in work of Bourgain and Lindenstrauss in connection with the QUE conjecture. In particular, from Theorem 5.1 of their paper it follows that there is a positive constant $\delta$ such that at least $p^{\delta}$ of the primes $\ell$ below $p$ are quadratic non-residues $\pmod p$. The proof is based on the fundamental lemma of sieve theory together with cancelation in character sums.

Added: To elaborate, Theorem 5.1 of Bourgain and Lindenstrauss's paper shows that if $N=p^{\beta}$ with $\beta \ge 1/4+\epsilon$ then there exists $\alpha>0$ such that ($\ell$ runs over primes below) $$ \sum_{N^{\alpha} <\ell < N; (\frac{\ell}{p}) = -1} \frac{1}{\ell} \ge \frac 12 -\epsilon. $$ In particular the number of primes $\ell$ with $(\frac{\ell}{p}) =-1$ is trivially at least $(1/2-\epsilon)N^{\alpha}$. Now use this with $N=p$, and we deduce the result mentioned above. I didn't check the details, but I think one can get a pretty decent value of $\delta$ above -- maybe even as big as $3/8$ (the level of distribution is like $p^{\frac 34}$ and the sifting limit should be $1/2$).

Let $\chi(n)$ denote the quadratic character modulo $p$ (so $\chi(n) = 1$ if $n$ is a quadratic residue modulo $p$, and $\chi(n)=-1$ if $n$ is a quadratic nonresidue modulo $p$). The difference between the number of primes that are quadratic residues and quadratic nonresidues is exactly $\sum_{\ell\lt p} \chi(\ell)$ where $\ell$ denotes a prime. One can deduce information about $\sum_{\ell\lt p} \chi(\ell)$ from information about $\sum_{\ell\lt p} \chi(\ell)\log\ell$, which in turn is almost the same as $\sum_{n\lt p} \chi(n)\Lambda(n)$ where $\Lambda$ is the von Mangoldt function. Such information is classically known, since the proof of the prime number theorem for arithmetic progressions hinges on it; it's important to note here that the summation goes up to $p$ itself rather than a general large $x$ as is typical for such statements. The answer then depends upon what zero-free region for the associated $L(s,\chi)$ you want to use or assume; if you get a bound that is $o(p)$, then the numbers of prime quadratic residues and prime quadratic nonresidues are very close to equal.

A comment to GH's elementary lower bound in question (1): Maybe there's a very minor error here. For example, if $p=7$ and $x=2$, then $x^2-p' = 11$ is not built up of primes enumerated in (1). But this is easily resolved by restricting to odd values of $x$ in the case when $p\equiv3\pmod{4}$, and noting we then already know the prime factor $2$ of $x^2-p'$.

More substantial comment: Doesn't this construction give something a bit better than $\log p/\log\log p$? If $q_1, \dots, q_k$ are the primes enumerated in (1), then we get that $\gg\sqrt{p}$ numbers in $[1, 2p]$ are supported on primes from the list $2, q_1, \dots, q_k$. But the count of numbers in $[1,2p]$ supported on this set of primes is at most the count of numbers supported on the primes $2, 3, 5, \dots, p_{k+1}$, where $p_i$ denotes the $i$th prime in increasing order. In other words, it's at most $\Psi(2p, p_{k+1})$. It is known from the theory of smooth numbers that $\Psi(x, (\log x)^{A}) = x^{1-1/A +o(1)}$, as $x\to\infty$. So it looks like GH's argument gives that $k \geq (\log{p})^{2-o(1)}$, as $p\to\infty$. (Of course this is a little less elementary.)