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.
A solution using Abel's summation as suggested by Cornel.
Let $\ \displaystyle S\ $ denote $\ \displaystyle \sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}\ $
and by using Abel's summation:
$\displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)\ $ where $\ \displaystyle A_n=\sum_{i=1}^n a_i\ $
and letting let $\ \displaystyle a_k=\frac{1}{(2k+1)^2}\ $ , $\ \displaystyle b_k=H_k^{(2)}$, we get
\begin{align}
\sum_{k=1}^n\frac{H_k^{(2)}}{(2k+1)^2}&=\sum_{i=1}^n\frac{H_{n+1}^{(2)}}{(2i+1)^2}-\sum_{k=1}^n\frac{1}{(k+1)^2}\left(\sum_{i=1}^k\frac{1}{(2i+1)^2}\right)\\
&=\sum_{i=1}^n\frac{H_{n+1}^{(2)}}{(2i+1)^2}-\sum_{k=1}^n\frac{1}{(k+1)^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}+\frac{1}{(2k+1)^2}-1\right)
\end{align}
Letting $n$ approach $\infty$, we get
\begin{align}
S&=\zeta(2)\sum_{i=1}^\infty\frac{1}{(2i+1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)\\
&\quad-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}+\sum_{k=1}^\infty\frac1{(k+1)^2}\\
&=\zeta(2)\left(\frac34\zeta(2)-1\right)-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}-\frac{1}{(2k-1)^2}\right)\\
&\quad-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}+\zeta(2)-1\\
&=\frac{15}8\zeta(4)-1-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)+\sum_{k=1}^\infty\frac{1}{k^2(2k-1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}\\
&=\frac{15}8\zeta(4)-1-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)+1\\
&\quad+\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}-\sum_{k=1}^\infty\frac{1}{(k+1)^2(2k+1)^2}\\
&=\frac{15}8\zeta(4)-\sum_{k=1}^\infty\frac{1}{k^2}\left(H_{2k}^{(2)}-\frac14H_{k}^{(2)}\right)\\
&=\frac{15}8\zeta(4)-4\sum_{k=1}^\infty\frac{H_{2k}^{(2)}}{(2k)^2}+\frac14\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}\\
&=\frac{15}8\zeta(4)-4\left(\frac12\sum_{k=1}^\infty\frac{H_{k}^{(2)}}{k^2}+\frac12\sum_{k=1}^\infty\frac{(-1)^kH_k^{(2)}}{k^2}\right)+\frac14\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}\\
&=\frac{15}8\zeta(4)-\frac74\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}-2\sum_{k=1}^\infty\frac{(-1)^kH_k^{(2)}}{k^2}
\end{align}
By plugging $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^nH_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42\ $
( proved here ) and $\ \displaystyle\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^2}=\frac74\zeta(4)\ $, we get the closed form of $\ S$
Best Answer
In the question body in Eq $(3)$, we reached
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2a}}=-\left(a+\frac12\right)\eta(2n+1)+\frac1{2(2a-1)!}\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx\tag{1}$$
From following the same approach of this solution, we have
\begin{align} I_a=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx&= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \frac{\partial}{\partial n} \operatorname{B}(m,n-m) \, \Bigg \rvert_{m=0, \, n=1} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \operatorname{\Gamma}(m) \frac{\partial}{\partial n} \frac{\operatorname{\Gamma}(n-m)}{\operatorname{\Gamma}(n)} \, \Bigg \rvert_{m=0,\, n=1} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \operatorname{\Gamma}(m) \operatorname{\Gamma}(1-m) [\operatorname{\psi}^{(0)} (1-m) + \gamma] ~\Bigg \rvert_{m=0} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}}\frac{\pi}{\sin(\pi m)} [\operatorname{\psi}^{(0)} (1-m) + \gamma] ~\Bigg \rvert_{m=0} \\ \end{align}
Edit: Thanks to @Gary for finding the closed form of this limit. I will mention it here with more details:
We have $$ \psi(1-z)+\gamma=\sum\limits_{n= 1}^\infty {\frac{1}{{n!}}\left[ {\frac{{d^n\psi(1-z)}}{{dz^n }}}\right]_{z=0}z^n}=\sum\limits_{n = 1}^\infty {(-1)^n \frac{{\psi^{(n)}(1)}}{{n!}}z^n} $$ for $|z|<1$. Thus, $$ \frac{{\psi(1-z)+\gamma}}{z}=\sum\limits_{n=0}^\infty {(-1)^{n+1} \frac{{\psi^{(n+1)}(1)}}{{(n+1)!}}z^n} $$ for $|z|<1$ (the left-hand side is defined as a limit when $z=0$). We also have $$ \pi z\csc(\pi z)=\sum\limits_{n=0}^\infty {\frac{{(-1)^{n-1}2\pi^{2n} (2^{2n -1}-1)B_{2n}}}{{(2n)!}}z^{2n}}=2\sum\limits_{n=0}^\infty {\eta (2n)z^{2n} } $$ for $|z|<1$ (the left-hand side is defined as a limit when $z=0$).
so we have $$F(z)=\pi\csc(\pi z)[\psi(1-z)+\gamma]=\left(2\sum\limits_{n = 0}^\infty {\eta (2n)z^{2n}}\right)\left(\sum\limits_{n=0}^\infty {(-1)^{n+1} \frac{{\psi^{(n+1)}(1)}}{{(n+1)!}}z^n}\right).$$ Apply Cauchy product:
$$\left(\sum_{n=0}^\infty a_{2n} z^{2n}\right)\left(\sum_{n=0}^\infty b_{n+1} z^n\right)=\sum_{n=0}^\infty\left(\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}a_{2k} b_{n-2k+1}\right)z^n$$
we get
$$F(z)=2\sum_{n=0}^\infty\left(\underbrace{\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor} \eta(2k) (-1)^{n-2k+1}\frac{\psi^{(n-2k+1)}}{(2-2k+1)!}}_{f_n}\right)z^n=2\sum_{n=0}^\infty f_n z^n.$$
Note that
\begin{align} \lim_{z\to0}\frac{\partial^{2a-1}}{\partial z^{2a-1}}F(z)&=2\lim_{z\to0}\frac{\partial^{2a-1}}{\partial z^{2a-1}} \sum_{n=0}^\infty f_n z^n\\ &=2\lim_{z\to0}\frac{\partial^{2a-1}}{\partial z^{2a-1}}\left(f_0 z^0+f_1 z^1+f_2 z^2+...\right)\\ &=2(2a-1)! f_{2a-1}\\ &=2(2a-1)!\sum_{k=0}^{\lfloor \frac{2a-1}{2}\rfloor} \eta(2k)(-1)^{2a-2k}\frac{\psi^{(2a-2k)}}{(2a-2k)!}\\ &=2(2a-1)!\sum_{k=0}^{a-1}\frac{\eta(2k)}{(2a-2k)!}\psi^{(2a-2k)}(1). \end{align} Substitute $\psi^{(a)}(1)=(-1)^{a-1}a!\zeta(a+1)$,
$$\lim_{z\to 0}\frac{\partial^{2a-1}}{\partial z^{2a-1}}F(z)=-2(2a-1)!\sum_{k=0}^{a-1}\eta(2k)\zeta(2a-2k+1)=-I_a.\tag{2}$$
Plug $(2)$ in $(1)$ we get
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\sum_{k=0}^{a-1}\eta(2k)\zeta(2a-2k+1).$$
acknowledgement:
Big thanks to @ComplexYetTrivial and @Gary for their solutions here and here. With their help, the proof is now completed rigorously which I think is a new proof in the literature.
More rigorous proof of $\displaystyle\pi z\csc(\pi z)=\sum_{n=0}^\infty \eta(2n) x^{2n}:$
Set $x=0$ in the Fourier series of $\cos(z x)$: $$\cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}-\sum_{n=1}^\infty\frac{(-1)^n\cos(nx)}{n^2-z^2}\right],$$ we get \begin{align} \pi z \csc(\pi z)&=1-2\sum_{n=1}^\infty\frac{(-1)^n z^2}{n^2-z^2}\\ &=1-2\sum_{n=1}^\infty (-1)^n \frac{z^2/n^2}{1-z^2/n^2}\\ &=1-2\sum_{n=1}^\infty (-1)^n \left(\sum_{k=1}^\infty (z^2/n^2)^k\right)\\ &=1-2\sum_{k=1}^\infty z^{2k} \left(\sum_{n=1}^\infty \frac{(-1)^n}{n^{2k}}\right)\\ &=1-2\sum_{k=1}^\infty z^{2k} \left(-\eta(2k)\right)\\ &=2\left(\frac12+\sum_{k=1}^\infty \eta(2k) z^{2k}\right)\\ &=2\sum_{k=0}^\infty \eta(2k) z^{2k}. \end{align}