For $p\equiv1\bmod4$,
we write
$$p=s_p^2+t_p^2=\sigma_p\overline{\sigma}_p,\ \ \ \sigma_p\in\mathbf{Z}[i].$$
Note that $\sigma_p$ and $2$ are coprime in $\mathbf{Z}[i].$ For each positive integer $k$, we would like to evaluate
$$ S_k(x):=\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x,\\ \sigma_p+\overline{\sigma}_p\equiv0\bmod4}}(\Re\sigma_p)^k
=\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x,\\ \sigma_p+\overline{\sigma}_p\equiv0\bmod4}}(\sqrt{p}\cos\theta_p)^k, $$
where $'$ yields the restriction that $\arg(\sigma_p)\in(0,\pi/2).$
On the other hand,
$$\sigma_p+\overline{\sigma}_p\equiv0\bmod4\Leftrightarrow \sigma_p^2+p\equiv0\bmod4\Leftrightarrow \sigma_p^2+1\equiv0\bmod4.$$
The last congruence is viewed $\mathbf{Z}[i]$, so that
$$S_k(x)=\frac{1}{4}\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x,\\ \sigma_p^2+1\equiv0\bmod4}}(\sqrt{p}\cos\theta_p)^k.$$
Introducing multiplicative characters in $(\mathbf{Z}[i]/4\mathbf{Z}[i])^\times$, we may write
$$ S_k(x)=\frac{1}{4}\frac{1}{\Phi(4)}\sum_{\chi\in\widehat{(\mathbf{Z}[i]/4\mathbf{Z}[i])^\times}}\sum_{z\in\mathbf{Z}[i],z^2+1\equiv0\bmod4}\overline{\chi}(z)\sideset{}{'}\sum_{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x}\chi(\sigma_p)(\sqrt{p}\cos\theta_p)^k,$$
where $\Phi(4)=|(\mathbf{Z}[i]/4\mathbf{Z}[i])^\times|=8.$
For each $\chi$, the sum
$$
\sideset{}{'}\sum_{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x}\chi(\sigma_p)(\cos\theta_p)^k,
$$
can be evaluated via Hecke's argument since
\begin{align*}
\chi(\sigma_p)e^{\ell i\theta_p}=\chi(\sigma_p)\Big(\frac{\sigma_p}{|\sigma_p|}\Big)^\ell
\end{align*}
gives a Hecke Grossencharacter at evaluated at $\sigma_p.$

Similarly, we can also consider
\begin{align*}
T_k(x):=\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x\\ \sigma_p-\overline{\sigma}_p\equiv0\bmod4}}(\Im\sigma_p)^k,
\end{align*}
which is related to the congruence $z^2-1\equiv0\bmod4.$

After a collection of serious arguments, we find the limit is equal to the ratio $|\mathcal{A}|/|\mathcal{B}|$
with
$$\mathcal{A}=\{z\bmod4:z^2+1\equiv0\bmod4\},\ \ \ \mathcal{B}=\{z\bmod4:z^2-1\equiv0\bmod4\}.$$
In fact,
\begin{align*}
\mathcal{A}=\{i,3i,2+i,2+3i\bmod4\},\ \ \ \mathcal{B}=\{1,1+i,3,3+2i\bmod4\}.
\end{align*}
Hence $|\mathcal{A}|/|\mathcal{B}|=1,$ which proves the first conjecture.

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