Did this for the case for $5$ is a quadratic residue.
That is, the problem equivalently says that we want to find all $p > 5$ for which $(\frac{5}{p}) = 1$. So using reciprocity, rewrite as $\left(\frac{p}{5} \right)$, which $ =1$ when $p$ is quadratic residue modulo $5$. So $p \equiv 1\pmod 5$ or $4 \pmod 5$.
Now for the case of $10$ a quadratic residue $mod\ p$. The problem equivalently says that we want to find all $p > 5$ for which $(\frac{10}{p}) = 1$. We can write $(\frac{10}{p}) = (\frac{5}{p})(\frac{2}{p})$, and the former, we know from above, and the latter $= 1$ when $p \equiv \pm 1 \pmod 8 $.
So to finish off the question, do I stitch together the solutions using C.R.T? Do I just say it's solutions where $p \equiv \pm 1 \pmod 8 $ and $p \equiv \pm 1 \pmod 5\ or\ 4 \pmod 5$?