Concerning this sum $\sum_{n=0}^{\infty}\frac{1}{4n+1}\left[\frac{1}{4^n}{2n \choose n}\right]^2=\frac{\Gamma^4\left(\frac{1}{4}\right)}{16\pi^2}$

From the beautiful answer by @robjohn, we learn \begin{equation} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac{r^{2k-1}}{(2k-1)^2} =\frac2{\pi r}\int_0^1\frac{\left(rx\arcsin(rx)+\sqrt{1-r^2x^2}\right)\,\mathrm{d}x}{\sqrt{1-x^2}} \end{equation} Plugging in $r=i$, \begin{align} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac{(-1)^k}{(2k-1)^2} &=\frac{2}{\pi}\int_0^1\frac{\left(ix\arcsin(ix)+\sqrt{1+x^2}\right)\,\mathrm{d}x}{\sqrt{1-x^2}}\\ &=\frac{2}{\pi}\int_0^1\frac{\sqrt{1+x^2}-x\sinh^{-1}(x)}{\sqrt{1-x^2}}\,\mathrm{d}x\\ &=\frac{2}{\pi}\int_0^1\left[ \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}-\frac{\sqrt{1-x^2}}{\sqrt{1+x^2}}\right]\,\mathrm{d}x\\ &=\frac{4}{\pi}\int_0^1 \frac{x^2}{\sqrt{1-x^4}}\,\mathrm{d}x\\ &=\frac{1}{\pi}B\left( \frac{3}{4},\frac{1}{2} \right)\\ &=\frac{4\sqrt{2\pi}}{\Gamma^2\left( \frac{1}{4} \right)} \end{align} (the integration of the $\sinh^{-1}$ term was made by parts).

For the second expression, we decompose \begin{equation} \frac{n^2}{\left( 2n-1 \right)^2}=\frac{1}{4}+\frac{1}{2}\frac{1}{2n-1}+\frac{1}{4}\frac{1}{\left( 2n-1 \right)^2} \end{equation} From the linked answer we have also \begin{equation} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac{r^{2k-1}}{2k-1} =\frac1{\pi r}\int_0^1\frac{-\sqrt{1-r^2x}\,\mathrm{d}x}{\sqrt{x(1-x)}} \end{equation} and thus, with $r=i$ \begin{align} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac{(-1)^k}{2k-1} &=-\frac1{\pi }\int_0^1\frac{\sqrt{1+x}\,\mathrm{d}x}{\sqrt{x(1-x)}}\\ &=-\frac2{\pi }\int_0^1\frac{\sqrt{1+t^2}}{\sqrt{1-t^2}}\,\mathrm{d}t\\ &=-\frac{2\sqrt{2}}{\pi}E\left( \frac{1}{\sqrt{2}} \right)\\ &=-\frac{2\sqrt{2}}{\pi}\left[ \frac{\Gamma^2\left( \frac{1}{4} \right)}{8\sqrt{\pi}}+\frac{\pi^{3/2}}{\Gamma^2\left( \frac{1}{4} \right)}\right] \end{align} where the singular value $k_1$ for the elliptic integral is used. We may also express from the answer \begin{equation} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}r^{2k} =\frac1\pi\int_0^1\frac{\mathrm{d}x}{\sqrt{1-r^2x}\sqrt{x(1-x)}} \end{equation} which, with $r=i$ again, gives \begin{align} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}(-1)^{k} &=\frac1\pi\int_0^1\frac{\mathrm{d}x}{\sqrt{1+x}\sqrt{x(1-x)}}\\ &=\frac2\pi\int_0^1\frac{dt}{\sqrt{1-t^4}}\\ &=\frac{1}{2\pi}B\left(\frac{1}{4}, \frac{1}{2} \right)\\ &=\frac{\sqrt{2\pi}}{2\Gamma^2\left(\frac{3}{4}\right)} \end{align} Then, \begin{align} S&=\sum_{n=0}^{\infty}\frac{(-1)^nn^2}{(2n-1)^2}\left[\frac{1}{4^n}{2n \choose n}\right]^2\\ &=\frac{\sqrt{2\pi}}{8\Gamma^2\left(\frac{3}{4}\right)} -\frac{\sqrt{2}}{\pi}\left[ \frac{\Gamma^2\left( \frac{1}{4} \right)}{8\sqrt{\pi}}+\frac{\pi^{3/2}}{\Gamma^2\left( \frac{1}{4} \right)}\right]+\frac{\sqrt{2\pi}}{\Gamma^2\left( \frac{1}{4} \right)} \end{align} Using the reflection formula $\Gamma(3/4)\Gamma(1/4)=\pi\sqrt{2}$, we obtain finally \begin{equation} S=-\frac{\Gamma^2\left( \frac{1}{4} \right)}{8\sqrt2\pi^{3/2}} \end{equation} as expected.

Edit (simpler derivation for the second expression)

For the second expression, it is easier to leave the integrals unevaluated in the decomposition: \begin{equation} S=\int_0^1\left[\frac{1}{2\pi}\frac{1}{\sqrt{1-x^4}}-\frac{1}{\pi}\sqrt{\frac{1+x^2}{1-x^2}}+\frac{1}{\pi}\frac{x^2}{\sqrt{1-x^4}} \right]\,dx \end{equation} or \begin{align} S&=-\frac{1}{2\pi}\int_0^1\frac{dx}{\sqrt{1-x^4}}\\ &=-\frac{1}{8\pi}\int_0^1 \left( 1-u \right)^{-1/2}u^{-3/4}\,du\\ &=-\frac{B\left( \frac{1}{4},\frac{1}{2} \right)}{8\pi} \end{align} and the result follows from the $\Gamma$ reflection formula.