Yes, to (1) and (2), but the argument below is much too crude to identify $c$.

The obvious attempt is to use $\sin t\ge 2t/\pi$, $0\le t\le \pi/2$ (or something similar), for the terms with large $n$. So we can set $N=N(x)=\lceil \sqrt{2x/\pi}\rceil$, and we obtain
$$
f(x)\ge \frac{2x}{\pi} \sum_{n=N}^{\infty} \frac{1}{n^2} - N+1\ge \frac{2x}{\pi (N+1)}-N+1\ge \frac{2x}{\sqrt{2\pi x}+2\pi}-\sqrt{\frac{2x}{\pi}} .
$$
This is asymptotically $(\sqrt{2/\pi}-\sqrt{2/\pi})\sqrt{x}$, so doesn't quite do it, but even the tiniest improvement will produce answers.

This is easy to do. For example, when $n\ge 2N$, then $x/n^2\le \pi/8$, and we obtain an improved bound $\sin x/n^2\ge cx/n^2$, with $c>2/\pi$ for these $n$. Since the part $\sum_{N\le n\le 2N} 1/n^2 \simeq 1/(2N)$ contributes only a fraction (one half, to be precise) of the full sum, the improvement of the constant on the other part will make itself felt in an improved overall constant.

This shows that $\liminf f(x)/\sqrt{x}>0$ (and gives an explicit but crude lower bound), and also that $f>0$ since one can keep track of the error terms and $x$ of moderate size has already been discussed by the OP (numerically).