The first observation follows from the law of quadratic reciprocity. Indeed, assume that $p\equiv 1\pmod{8}$ and $p=A^2+B^2$. Let $A'$ denote the odd part of $A$. Then $p\equiv B^2\pmod{A'}$, and hence $\left(\frac{p}{A'}\right)=1$. It follows that $\left(\frac{A'}{p}\right)=1$, and then by $\left(\frac{2}{p}\right)=1$ also $\left(\frac{A}{p}\right)=1$. In the same way, $\left(\frac{B}{p}\right)=1$.
The second observation lies deeper. I will use Gerry Meyerson's helpful comment below the original post. For any prime $p=4n+1$, Gauss (1825) observed that $p=A^2+B^2$, where $A$ and $B$ are the absolute least residues modulo $p$ of $(2n)!/(2n!^2)$ and $(2n)!^2/(2n!^2)$. For a proof, see Jacobsthal (1907) or Barnes (1974). Hence it suffices to show that in case when $4$ divides $n$, these two integers are fourth powers modulo $p$. So let us assume that $4\mid n$. By Wilson's theorem, $$(2n)!^{(p-1)/4}=(2n)!^{n}\equiv(p-1)!^{n/2}\equiv(-1)^{n/2}=1\pmod{p}.$$ Therefore, $(2n)!$ is a fourth power modulo $p$, and we only need to prove that $2n!^2$ is a fourth power modulo $p$. This is equivalent to $$2^{(p-1)/4}n!^{(p-1)/2}\equiv 1\pmod{p}.$$ Now $2$ is a square modulo $p$, hence $2^{(p-1)/4}\equiv\pm 1\pmod{p}$ depending on whether $2$ is a fourth power modulo $p$ or not. So we need to show that $$\left(\frac{n!}{p}\right)=\left(\frac{2}{p}\right)_4.$$ However, this follows from Theorem 5 in Emma Lehmer's 1977 paper "Generalizations of Gauss's lemma" applied to the $n$ quadratic non-residues lying in $\{1,\dots,2n\}$, and we are done.