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 $$S_k(x)=\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x,\\ \sigma_p^2+1\equiv0\bmod4}}(\sqrt{p}\cos\theta_p)^k.$$$$S_k(x)=\frac{1}{4}\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x,\\ \sigma_p^2+1\equiv0\bmod4}}(\sqrt{p}\cos\theta_p)^k.$$ 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 seriesserious arguments, we may arrive atfind 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.$$|\mathcal{A}|/|\mathcal{B}|=1,$ which proves the first conjecture.