I am going to establish two relations as Cornel suggested and solve them by elimination.
From here we have
$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$
Multiply both sides by $\frac{H_{2n}}{4n}$ then sum them from $n=1$ to $\infty$ we get
\begin{align}
R_1&=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}\left(H_n^2+H_n^{(2)}\right)=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{x^nH_{2n}}{2n}\ dx\quad \color{red}{x\mapsto x^2}\\
&=\int_0^1\frac{\ln^2(1-x^2)}{x}\sum_{n=1}^\infty\frac{x^{2n}H_{2n}}{2n}\ dx, \quad \color{red}{2\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{n}+\sum_{n=1}^\infty (-1)^na_{n}}\\
&=\frac12\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\sum_{n=1}^\infty\frac{x^{n}H_{n}}{n}+\sum_{n=1}^\infty\frac{(-x)^{n}H_{n}}{n}\right)\ dx\\
&=\frac12\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\underbrace{\operatorname{Li}_2(x)+\operatorname{Li}_2(-x)}_{\frac12\operatorname{Li}_2(x^2)}+\frac12\ln^2(1-x)+\frac12\ln^2(1+x)\right)\ dx\\
&=\small{\frac14\underbrace{\int_0^1\frac{\ln^2(1-x^2)\operatorname{Li}_2(x^2)}{x}\ dx}_{x^2\mapsto x}+\frac14\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx}\\
&=\frac18\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x)}{x}\ dx+\frac14\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx\\
&=\frac18A+\frac14B\tag{1}
\end{align}
Let's start with $A$ and by seting $1-x\mapsto x$ we can write
\begin{align}
A&=\int_0^1\frac{\ln^2x\color{blue}{\operatorname{Li}_2(1-x)}}{1-x}\ dx\\
&=\int_0^1\frac{\ln^2x}{1-x}\left[\color{blue}{\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)}\right]\ dx\\
&=\zeta(2)\int_0^1\frac{\ln^2x}{1-x}\ dx-\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx-\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\ dx-\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^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{A=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 $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$
To evaluate $B$, we are going to use the key identity
$$(a+b)^2(a^2+b^2)=\frac23a^4+\frac23b^4+\frac5{12}(a+b)^4-\frac1{12}(a-b)^4$$
and by taking $a=\ln(1-x)$ and $b=\ln(1+x)$ , we get
$$B=\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx\\
=\small{\frac23\int_0^1\frac{\ln^4(1-x)}{x}\ dx+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac5{12}\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}-\frac1{12}\underbrace{\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}}\\
=\frac78\int_0^1\frac{\ln^4(1-x)}{x}\ dx+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx-\frac16\int_0^1\frac{\ln^4x}{1-x^2}\ dx\\
=\frac78(24\zeta(5))+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx-\frac16\left(\frac{93}{4}\zeta(5)\right)\\
=\frac{137}{8}\zeta(5)+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx$$
Since
\begin{align}
\int_0^1\frac{\ln^4(1+x)}{x}&=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\
&=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\
&=\frac15\ln^52+\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\
&=\small{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)}
\end{align}
Then
$$\boxed{B=\frac83\ln^32\zeta(2)-7\ln^22\zeta(3)+\frac{265}{8}\zeta(5)-\frac8{15}\ln^52-16\ln2\operatorname{Li}_4\left(\frac12\right)-16\operatorname{Li}_5\left(\frac12\right)}$$
Plugging the boxed results of $A$ and $B$ in (1) we get our first relation:
$$R_1=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}\left(H_n^2+H_n^{(2)}\right)\\
=\small{\frac23\ln^32\zeta(2)-\frac74\ln^22\zeta(3)+\frac14\zeta(2)\zeta(3)+\frac{261}{32}\zeta(5)-\frac2{15}\ln^52-4\ln2\operatorname{Li}_4\left(\frac12\right)-4\operatorname{Li}_5\left(\frac12\right)}$$
We have
$$\frac{\ln^2(1-y)}{1-y}=\sum_{n=1}^\infty y^n(H_n^2-H_n^{(2)})$$
integrate both sides from $y=0$ to $y=x$ to get
$$-\frac13\ln^3(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^2-H_n^{(2)}\right)=\sum_{n=1}^\infty\frac{x^{n}}{n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$
Now replace $x$ with $x^2$ then multiply both sides by $-\frac{\ln(1-x)}{x}$ and integrate from $x=0$ to $x=1$, also note that $\int_0^1 -x^{2n-1}\ln(1-x)\ dx=\frac{H_{2n}}{2n}$ we get
$$\frac13\underbrace{\int_0^1\frac{\ln^3(1-x^2)\ln(1-x)}{x}\ dx}_{\large C}=\sum_{n=1}^\infty\frac{H_{2n}}{2n^2}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$
Rearranging the terms to get
$$R_2=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}(H_n^2-H_n^{(2)})=4\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n)^3}-8\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}+\frac16C\tag{2}$$
Cornel elegantly calculated the first sum here
$$\boxed{\small{\sum _{n=1}^{\infty } \frac{H_{2 n}H_n }{(2 n)^3}=\frac{307}{128}\zeta(5)-\frac{1}{16}\zeta (2) \zeta (3)+\frac{1}{3}\ln ^3(2)\zeta (2) -\frac{7}{8} \ln ^2(2)\zeta (3)-\frac{1}{15} \ln ^5(2)
-2 \ln (2) \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right)}}$$
For the second sum:
$$\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac12\sum_{n=1}^\infty\frac{H_{n}}{n^4}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}$$
plugging the common results:
$$\sum_{n=1}^\infty\frac{H_{n}}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$
$$\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$
we get
$$\boxed{\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac{37}{64}\zeta(5)-\frac14\zeta(2)\zeta(3)}$$
For the remaining integral $C$, we use the magical identity
$$(a+b)^3a=a^4-b^4+\frac12(a+b)^4-\frac12(a-b)^4-(a-b)^3b$$
with $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write
$$C=\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac12\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}\\-\underbrace{\frac12\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}-\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}$$
$$C=\frac54\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{4!\zeta(5)}-\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{K}-\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}{4}\zeta(5)}+\underbrace{2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx}_{J}$$
$$C=\frac{27}{4}\zeta(5)-K+J\tag{3}$$
we have already evaluated $K$ above:
$$K=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)$$
for $J$
$$J=2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx=\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx+\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1+x}\ dx$$
using the rule
$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{a+1}}{n}$$
allows us to write
\begin{align}
J&=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}\ dx-\ln2\int_0^1\frac{\ln^3x}{1+x}\ dx\\
&=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-\sum_{n=1}^\infty(-1)^n H_n\int_0^1x^n\ln^3x\ dx-\ln2\left(-\frac{21}4\zeta(4)\right)\\
&=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+6\sum_{n=1}^\infty\frac{(-1)^n H_n}{(n+1)^4}+\frac{21}{4}\ln2 \zeta(4)\\
&=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-6\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^4}-\frac{45}{8}\zeta(5)+\frac{21}{4}\ln2 \zeta(4)
\end{align}
Plugging
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}=\frac78\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)-2\zeta(5)$$
we get
$$J=\frac{279}{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)$$
Plugging the results of $K$ and $J$ in (3) we get
$$\boxed{\small{C=24\operatorname{Li}_5\left(\frac12\right)+24\ln2\operatorname{Li}_4\left(\frac12\right)+\frac3{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)+\frac{21}2\ln^22\zeta(3)-4\ln^32\zeta(2)+\frac45\ln^52}}$$
and by substituting the boxed results in (2) we get our second relation
$$R_2=\sum _{n=1}^{\infty } \frac{H_{2 n} }{(2 n)^2}(H_n^2-H_n^{(2)})
=\frac78\zeta(2)\zeta(3)+5\zeta (5)+\frac{2}{3} \ln ^3(2)\zeta (2) -\frac{7}{4} \ln ^2(2)\zeta (3)\\ -\frac{2}{15} \ln^5(2)
-4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$
Thus
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{R_1-R_2}{2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3)$$
and as a bonus
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^2}{(2n)^2}=\frac{R_1+R_2}{2}\\
=\small{\frac{421 }{64}\zeta (5)+\frac{9 }{16}\zeta (2) \zeta (3)+\frac{2}{3} \ln ^32\zeta (2) -\frac{7}{4} \ln ^22\zeta (3) -\frac{2}{15} \ln^52-4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right)}$$
Note:
$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ can be found here and $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}$ can be found here.