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
Here is another solution as requested.
§1. The sum
$$s_3=\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$$
has two aspects of interest (a) the structure of the demoniminator as a (rising) Pochhammer symbol and (b) the quadratic expression in harmonic numbers in the summand.
As far as I know this is an original problem which was proposed and solved for the first time in the context of the previously deleted question. Please correct me if I'm wrong.
§2. My approach reduces the sum in question to a linear combination of sums over summands of the type $1/p(n)$ and $H/p(n)$ where $H$ is a (generalized) harmonic number and $p(n)$ is a polynomial.
To begin with, let us look the simpler more general sum
$$a_2=\sum_{n=1}^\infty\frac{a(n)}{(n+1)(n+2)}$$
We transform the summand as follows: partial fraction decomposition gives
$$\frac{a(n)}{(n+1)(n+2)}=\frac{a(n)}{1+n}-\frac{a(n)}{2+n}$$
Now we add and subtract a term which is the first term with the index shifted upwards by 1, i.e.
$$\frac{a(n)}{1+n}|_{n\to n+1} = \frac{a(n+1)}{2+n}$$
with the intention of creating terms that telescope under the sum.
Hence
$$\frac{a(n)}{1+n}-\frac{a(n)}{2+n}=\left(\frac{a(n)}{1+n}-\frac{a(n+1)}{2+n}\right) +\left(\frac{a(n+1)}{2+n}-\frac{a(n)}{2+n}\right)$$
we see that the first bracket telescopes term under a sum. The remaining term contains the sequence of the differences of the original seqence.
Taking the finite sum leads to
$$\sum_{n=1}^{m}\frac{a(n)}{(n+1)(n+2)}\\ = \sum_{n=1}^{m}\left(\frac{a(n)}{1+n}-\frac{a(n+1)}{2+n}\right)+\sum_{n=1}^{m}\left(\frac{a(n+1)}{2+n}-\frac{a(n)}{2+n}\right)\\ =\left(\frac{a(1)}{2}-\frac{a(n+1)}{2+n}\right)+\sum_{n=1}^{m}\left(\frac{a(n+1)-a(n)}{2+n}\right)$$
Under the assumption that $\lim_{n\to \infty} {(a(n)/n)}=0$ the corresponding infinite sum reads
$$\sum_{n=1}^{\infty}\frac{a(n)}{(n+1)(n+2)}=\frac{a(1)}{2}+\sum_{n=1}^{\infty}\left(\frac{a(n+1)-a(n)}{2+n}\right)$$
§3. The case $s_3$ can be traced back to $s_2$ setting $a(n)\to \frac{a(n)}{n+3}$.
Applying the same procedure as before we arrive at
$$\sum_{n=1}^{\infty} \frac{a(n)}{(1+n)(2+n)(3+n)} = S_0+S_1+S_2$$
where
$$S_0 = -\frac{5}{25}a(1)+\frac{1}{6}a(2)+\frac{1}{8}a(3)$$ $$S_1 = \frac{1}{2}\sum_{n=1}^{\infty}\frac{a(n+3)-a(n+1)}{n+4}$$ $$S_2 = \sum_{n=1}^{\infty}\frac{a(n+1)-a(n)}{n+3}$$
Specifically for $a(n) = H_n H_n^{(2)}$ and observing that
$$H_{k+1} = H_{k}+\frac{1}{k+1}$$ $$H_{k+1}^{(2)}=H_{k}^{(2)}+\frac{1}{(k+1)^2}$$
we get the following combination of linearized sums, together with their exact sums (based on known results):
$$A_0=\frac{719}{1728}$$
$$\begin{align} A_1& = \sum _{n=1}^{\infty } \left(\frac{H_{n+1}}{2 (n+2)^2 (n+4)}+\frac{H_{n+1}}{2 (n+3)^2 (n+4)}\\ +\frac{H_{n+1}^{(2)}}{2 (n+2) (n+4)}+\frac{H_{n+1}^{(2)}}{2 (n+3) (n+4)}\\ +\frac{1}{2 (n+2)^3 (n+4)}+\frac{1}{2 (n+2)^2 (n+3) (n+4)}\\ +\frac{1}{2 (n+2) (n+3)^2 (n+4)}+\frac{1}{2 (n+3)^3 (n+4)}\right)\\ & =\frac{3 \zeta (3)}{2}+\frac{\pi ^2}{24}-\frac{2969}{1728} \end {align}$$
$$\begin{align} A_2& =\sum _{n=1}^{\infty } \left(-\frac{H_n}{(n+1)^2 (n+3)}-\frac{H_n^{(2)}}{(n+1) (n+3)}-\frac{1}{(n+1)^3 (n+3)}\right)\\ &=-\zeta (3)-\frac{\pi ^2}{12}+\frac{61}{48} \end{align}$$
Adding the three expressions we obtain
$$s_3 = \frac{\zeta (3)}{2}-\frac{\pi ^2}{24}-\frac{1}{32}$$
which is equivalent to the expression obtained in the OP since $\zeta(2) = \frac{\pi^2}{6}$.
$4. Generalization
The obvious generalization asks for the value of
$$s_q(a,b) = \sum_{n=1}^{\infty} \frac{H_n^{(a)} H_n^{(b)}}{(n+1)_q}$$
where $a$ and $b$ can assume the values $0$ or $1$ and
$$H_n^{(a)}=\sum_{k=1}^{n}\frac{1}{k^a}$$
is the generalized harmonic number and
$$(n+1)_q = (n+1)(n+2)...(n+q)$$
is the ascending Pochhammer symbol.
By the procedure described in this answer we can reduce the quadratic sum to a combination of linear sums which can be evaluated.