To calculate these two sums, we are going to establish two relations and solve them by elimination.
To establish the first relation, we use $\displaystyle I=\int_0^1\frac{\ln^4(1+x)+6\ln^2(1-x)\ln^2(1+x)}{x}\ dx=\frac{21}4\zeta(5)\tag{1}$
which was proved by Khalef Ruhemi ( unfortunately he is not an MSE user).
The proof as follows: using the algebraic identity $\ b^4+6a^2b^2=\frac12(a-b)^4+\frac12(a+b)^4-a^4$
with $\ a=\ln(1-x)$ and $\ b=\ln(1+x)$ , divide both sides by $x$ then integrate, we get
$$I=\frac12\underbrace{\int_0^1\frac1x{\ln^4\left(\frac{1-x}{1+x}\right)}\ dx}_{\frac{1-x}{1+x}=y}+\underbrace{\frac12\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2=y}-\int_0^1\frac{\ln^4(1-x)}{x}\ dx$$
$$=\int_0^1\frac{\ln^4x}{1-x^2}+\frac14\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1-x)}{x}\ dx$$
$$=\frac12\int_0^1\frac{\ln^4x}{1-x}+\frac12\int_0^1\frac{\ln^4x}{1+x}-\frac34\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{1-x=y}$$
$$=\frac12\int_0^1\frac{\ln^4x}{1+x}\ dx+\frac14\int_0^1\frac{\ln^4x}{1-x}\ dx=\frac12\left(\frac{45}{2}\zeta(5)\right)+\frac14(24\zeta(5))=\frac{21}4\zeta(5)$$
On the other hand, $\quad\displaystyle I=\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{I_1}+6\int_0^1\frac{\ln^2(1-x)\ln^2(1+x)}{x}\ dx$
Using $\ln^2(1+x)=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n\ $ for the second integral, we get
\begin{align}
I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln^2(1-x)\ dx\\
I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(\frac{H_n^2+H_n^{(2)}}{n}\right)\\
I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n^3+H_nH_n^{(2)}}{n^2}\right)-12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n^2+H_n^{(2)}}{n^3}\right)\tag{2}
\end{align}
From $(1)$ and $(2)$, we get
$$\boxed{\small{R_1=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\frac{7}{16}\zeta(5)+\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-\frac{1}{12}I_1}}$$
and the first relation is established.
To get the second relation, we need to use the sterling number formula ( check here)
$$ \frac{\ln^k(1-x)}{k!}=\sum_{n=k}^\infty(-1)^k \begin{bmatrix} n \\ k \end{bmatrix}\frac{x^n}{n!}$$
letting $k=4$ and using $\displaystyle\begin{bmatrix} n \\ 4 \end{bmatrix}=\frac{1}{3!}(n-1)!\left[\left(H_{n-1}\right)^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right],$ we get $$\frac14\ln^4(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
differentiate both sides with respect to $x$, we get
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
Now replace $x$ with $-x$ then multiply both sides by $\frac{\ln x}{x}$ and integrate, we get
$$-\sum_{n=1}^\infty(-1)^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1x^{n-1}\ln x\ dx=\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}\ dx$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)=\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx-\underbrace{\int_0^1\frac{\ln^3(1+x)\ln x}{1+x}\ dx}_{IBP}$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)=\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx+\frac14I_1$$
Rearranging the terms, we get
$$\boxed{R_2=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}-3\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\int_0^1\frac{\ln^3(1+x)\ln x}{x}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}+\frac14I_1}$$
and the second relation is established.
Now we are ready to calculate the first sum.
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}&=\frac{3R_1+R_2}{4}\\
&=\frac34\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\frac34\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}\\
&\quad+\frac14\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx+\frac{21}{64}\zeta(5)
\end{align}
the closed form of the first and second sum can be found here and the closed form of the third sum is evaluated here. as for the integral, I evaluated it here.
by combining these results, we get our closed form.
and the second sum.
$$\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\frac{R_1-R_2}{4}$$
$$\small{=\frac14\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\frac14\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}+\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}-\frac14\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx-\frac1{12}I_1+\frac{7}{64}\zeta(5)}$$
lets calculate $I_1$ and by setting $\frac1{1+x}=y$, we get
\begin{align}
I_1&=\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)\\
&=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}
by combining the result of $I_1$ along with the results we used in our first sum, we get the closed form of the second sum.
UPDATE:
The identity used above:
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
can also be proved this way.
We proved in this solution that
$$\mathcal{I}=\int_0^1\frac{\tan^{-1}(x)\ln(1+x^2)}{x(1+x)}dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^2(2)\tag1$$
and we proved here that
$$\tan^{-1}x\ln(1+x^2)=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}\tag2$$
By $(2)$ we get
$$\mathcal{I}=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1\frac{ x^{2n}}{1+x}dx$$
Using the identity $$\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$$
it follows that
$$\mathcal{I}=-2\ln(2)\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}+2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{2}}{2n+1}$$
$$=-2\ln(2)\mathcal{S}_1-2\mathcal{S}_2+2\mathcal{S}_3\tag3$$
For $\mathcal{S}_1$ and $\mathcal{S}_3$, we use the classical identity:
$$\sum_{n=1}^\infty(-1)^n f(2n)=\Re\sum_{n=1}^\infty i^n f(n)$$
Therefore
$$\mathcal{S}_1=\Re\sum_{n=1}^\infty i^n\frac{H_n}{n+1}=\Re\left\{\frac{\ln^2(1-i)}{i}\right\}=-\frac{\pi}{8}\ln(2)\tag4$$
where we used $\sum_{n=1}^\infty x^n\frac{H_n}{n+1}=\frac{\ln^2(1-x)}{x}$ which follows from integrating $\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$.
Similarly,
$$\mathcal{S}_3=\Re\sum_{n=1}^\infty i^n\frac{H_n^2}{n+1}$$
Using the generating function
$$\sum_{n=1}^\infty x^{n}\frac{ H_n^{2}}{n+1}=\frac{6\operatorname{Li}_3(1-x)-3\operatorname{Li}_2(1-x)\ln(1-x)-\ln^3(1-x)-3\zeta(2)\ln(1-x)-6\zeta(3)}{3x}$$
it follows that
$$\mathcal{S}_3=\Re\left\{\frac{6\operatorname{Li}_3(1-i)-3\operatorname{Li}_2(1-i)\ln(1-i)-\ln^3(1-i)-3\zeta(2)\ln(1-i)-6\zeta(3)}{3i}\right\}$$
$$=\Im\left\{\frac{6\operatorname{Li}_3(1-i)-3\operatorname{Li}_2(1-i)\ln(1-i)-\ln^3(1-i)-3\zeta(2)\ln(1-i)-6\zeta(3)}{3}\right\}$$
$$=2\Im\{\operatorname{Li}_3(1-i)\}+\frac12\ln(2)\ G+\frac{3\pi}{16}\ln^2(2)+\frac{5}{96}\pi^3\tag5$$
Plug the results of $(4)$ and $(5)$ in $(3)$ we get
$$\mathcal{I}=4\Im\{\operatorname{Li}_3(1-i)\}+\ln(2)\ G+\frac{5\pi}{8}\ln^2(2)+\frac{5}{48}\pi^3-2\mathcal{S}_2\tag6$$
By $(1)$ and $(6)$ we get
$$\mathcal{S}_2=\sum_{n=1}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}=2\Im\{\operatorname{Li}_3(1-i)\}+\frac12\ln(2)\ G+\frac{3\pi}{8}\ln^2(2)+\frac{3}{64}\pi^3$$
Best Answer
A fast and simple solution idea by Cornel Ioan Valean
We will use the power of the ideas and strategies from the books (Almost) Impossible Integrals, Sums, and Series (2019) and More (Almost) Impossible Integrals, Sums, and Series (2023).
Okay, let's start!
From the book, More (Almost) Impossible Integrals, Sums, and Series (2023), Sect. $4.5$, pages $396$-$398$, we have that $$\displaystyle \color{blue}{-\frac{\log(1-x^2)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}x^{2n} \frac{1}{4^n}\binom{2n}{n}(2 H_{2n}-H_n),\ |x|<1} ,$$ and exploiting this fact and turning the left-hand side into a double integral, $\displaystyle \int _0^{\pi/2}\left(\int _0^{\sin (x)}\frac{\log \left(1-y^2\right)}{y\sqrt{1-y^2}}\textrm{d}y\right)\textrm{d}x$, immediately reveals (after changing the integration order) that $$\sum _{n=1}^{\infty } \binom{2 n}{n}^2 \frac{2 H_{2 n}-H_n}{n 2^{4 n}}$$ $$ =\frac{\pi^2}{2}+\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1-x)}{x} \textrm{d}x+\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1+x)}{x} \textrm{d}x$$ $$-\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1+x^2)}{x} \textrm{d}x. \tag1$$
On the other hand, exploiting that $$\int_0^{\pi/2} \log(\sin(x)) \sin^{2n}(x) \textrm{d}x=\frac{\pi}{2}\frac{1}{2^{2n}}\binom{2n}{n}\left(H_{2n}-H_n-\log(2)\right),$$ which is also found in the sequel, page $191$, multiplying both sides by $\displaystyle \frac{1}{n 2^{2n}}\binom{2n}{n}$, making the summation from $n=1$ to $\infty$, and rearranging, we get that $$\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{H_{2n}-H_n-\log(2)}{n 2^{4n}}$$ $$\small =-\frac{8 }{\pi }\int_0^1 \frac{\log ^2(1+x^2)}{1+x^2} \textrm{d}x+\frac{8 }{\pi } \log (2)\int_0^1 \frac{ \log(1+x^2)}{1+x^2}\textrm{d}x$$ $$+\frac{8 }{\pi }\int_0^1 \frac{\log(x) \log(1+x^2)}{1+x^2} \textrm{d}x.\tag2$$
The rest is known and trivial, and the separate series involving the numbers $H_{2n}$ and $H_n$ are extracted by a system of relation consisting of $(1)$ and $(2)$.
End of story
Thank you so much Cornel for your work and your life-changing books, (Almost) Impossible Integrals, Sums, and Series (2019) and More (Almost) Impossible Integrals, Sums, and Series (2023).
(Very) important: Exploiting such techniques involving the series in blue for building systems of relations, one can derive extremely difficult series. Some (simple) paper(s) showing such examples will be written soon.
Update 1: Indeed, the series taken apart also contain a Trilogarithm with a complex argument. The forms of the two key identities (that is, using integrals) avoid that appearance, and to get directly the desired value without touching the Trilogarithm with a complex argument, all we need is to turn the integrals from the identity in $(2)$ into arctan-log integrals. Then we need to exploit the arctan-log integral identities given in the sequel, Sects. $1.36$-$1.38$, pages $48$-$51$. That's all (and done)!
Update 2: A full solution with all the details of this problem will be found soon in a separate paper (the main series plus the two separated series).
Update 3: Binoharmonic Series with the Squared Central Binomial Coefficient And Their Integral Transformation Using Elliptic Integrals by Cornel Ioan Valean