How to prove:
$$S_1=\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n)^2} =\frac{23 }{32}\zeta (2) \zeta (3)-\frac{581}{128} \zeta (5)-\frac{2}{3}\ln ^32 \zeta (2)+\frac{7}{4} \ln^22\zeta (3)\\ +\frac{2}{15} \ln ^52 +4\ln2 \operatorname{Li}_4\left(\frac{1}{2}\right) +4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$
$$S_2=\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n)^2} =\frac{23 }{32}\zeta (2) \zeta (3)+\frac{917 }{128}\zeta (5)+\frac{2}{3} \ln ^32\zeta (2)-\frac{7}{4} \ln ^22\zeta (3)\\-\frac{2}{15} \ln ^52 -4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right)-4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$
where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}, \ m\ge1,$ represents the $n$th generalized harmonic number of order $m$, $\zeta$ is the Riemann zeta function and $\operatorname{Li}_n$ is the polylogarithm function.
These two sums were proposed by Cornel here where he asked if these results and others ( in the link) exist in the literature.
I managed to find a relation between $S_1$ and $S_2$ so we need to find another relation or evaluate of them separately.
Here is how I got the relation:
From here we have
$$\int_0^1 x^{2n-1}\ln^2(1-x)\ dx=\frac{H_{2n}^2+H_{2n}^{(2)}}{2n}$$
multiply both sides by $\frac{H_n}{2n}$ then sum them from $n=1$ to $\infty$ to get
$$\sum_{n=1}^\infty\frac{H_n}{(2n)^2}(H_{2n}^2+H_{2n}^{(2)})=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty(x^2)^n\frac{H_n}{n}\ dx\\=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\left(\operatorname{Li}_2(x^2)+\frac12\ln^2(1-x^2)\right)\ dx\\=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\left(2\operatorname{Li}_2(x)+2\operatorname{Li}_2(-x)+\frac12\ln^2(1-x^2)\right)\ dx\\=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x)}{x}+\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(-x)}{x}+\frac14\int_0^1\frac{\ln^2(1-x)\ln^2(1-x^2)}{x}\ dx\\=A+B+\frac14C$$
\begin{align} A&=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x)}{x}\ dx=\int_0^1\frac{\ln^2x\operatorname{Li}_2(1-x)}{1-x}\ dx\\ &=\zeta(2)\int_0^1\frac{\ln^2x}{1-x}-\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}-\int_0^1\frac{\ln^2x\operatorname{Li}_2(x)}{1-x}\ dx\\ &=2\zeta(2)\zeta(3)+\sum_{n=1}^\infty H_n\int_0^1 x^n\ln^3x-\sum_{n=1}^\infty H_n^{(2)}\int_0^1x^n\ln^2x\ dx\\ &=2\zeta(2)\zeta(3)-6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}\\ &=2\zeta(2)\zeta(3)-6\sum_{n=1}^\infty\frac{H_n}{n^4}+6\zeta(5)-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+2\zeta(5)\\ &=\boxed{2\zeta(2)\zeta(3)-\zeta(5)} \end{align}
where we used $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ and $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$
\begin{align} B&=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(-x)}{x}\ dx\\ &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1x^{n-1} \ln^2(1-x)\ dx\\ &=\sum_{n=1}^\infty(-1)^n\frac{H_n^2+H_n^{(2)}}{n^3}\\ &=\boxed{\small{\frac23\ln^32\zeta(2)-\frac74\ln^22\zeta(3)+\frac{3}4\zeta(2)\zeta(3)+\frac{15}{16}\zeta(5)-\frac2{15}\ln^52-4\ln2\operatorname{Li}_4\left(\frac12\right)-4\operatorname{Li}_5\left(\frac12\right)}} \end{align}
where the results $$\small{\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}=\frac23\ln^32\zeta(2)-\frac74\ln^22\zeta(3)+\frac{3}4\zeta(2)\zeta(3)+\frac{15}{16}\zeta(5)-\frac2{15}\ln^52-4\ln2\operatorname{Li}_4\left(\frac12\right)-4\operatorname{Li}_5\left(\frac12\right)}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}=\frac{11}{32}\zeta(5)-\frac58\zeta(2)\zeta(3)$$
were used. Both series can be found here.
To find $C$, we are going to use the algebraic identity:
$$a^2(a+b)^2=\frac43a^4-\frac23b^4+\frac5{24}(a+b)^4+\frac{13}{24}(a-b)^4-(a-b)^3b$$
with $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write
$$C=\frac43\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac5{24}\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}-\frac{13}{24}\underbrace{\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}-\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}\\C=\small{\frac{23}{16}\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{24\zeta(5)}-\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx-\frac{13}{12}\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}4\zeta(5)}}-2\int_0^1\frac{\ln^2x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx\\C=\frac{149}{16}\zeta(5)-\frac23\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{K}+\underbrace{2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx}_{J}$$
The integrals $K$ and $J$ are calculated here
$$K=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)$$
$$J=\frac{279}{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)$$
Combining $K$ and $J$ gives
$$\boxed{\small{C=\frac{43}{4}\zeta(5)-\frac{21}4\zeta(2)\zeta(3)-\frac83\ln^32\zeta(2)+7\ln^22\zeta(3)+\frac8{15}\ln^52+16\ln2\operatorname{Li}_4\left(\frac12\right)+16\operatorname{Li}_5\left(\frac12\right)}}$$
now combine the boxed results of the integrals $A$, $B$ and $C$ we get
$$\sum_{n=1}^\infty\frac{H_n}{(2n)^2}(H_{2n}^2+H_{2n}^{(2)})=\frac{23}{16}\zeta (2) \zeta (3)+\frac{21}{8} \zeta (5)$$
