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.
Considering the algebraic identity
\begin{align*}
&(a-b)^3b = a^3b - 3a^2b^2 + 3ab^3 - b^4 = -2a^3b +3(a^3b+ab^3) -3a^2b^2 -b^4\\
&\Longrightarrow \ \ \ 2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b
\end{align*} with $a = \ln(1-x)$ and $b= \ln (1+x)$ it follows that
\begin{align*}
2\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =& - \frac 1 2\int_0^1 {\ln^4(1+x)\over x}d x \\
&-\frac 12 \int_0^1 \frac{\ln^4(1+x) + 6\ln^2(1-x)\ln^2(1+x)}{x}dx\\
&+3\int_0^1 \frac{\ln^3(1-x)\ln(1+x) + \ln(1-x)\ln^3(1+x)}{x}dx\\
&- \int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}dx\\
=:& -I_1 - I_2 + I_3 -I_4.
\end{align*}
For $I_1$, make substitution $y = \frac x {1+x}$ to get:
\begin{align*}
I_1 =& \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{y(1-y)} dy \\
=& \frac 1 2\underbrace{ \int_0^{\frac 12} \frac{\ln^4(1-y)}{y} dy}_{z=1-y}+ \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{1-y} dy\\
=& \frac 1 2 \int_{\frac 1 2 }^1 \frac{\ln^4 z} {1-z} dz + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \int_{\frac 1 2}^1 z^{n-1}\ln^4 z\ dz + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \frac{\partial^4}{\partial n^4}\left[\frac 1 n - \frac 1 {n2^n}\right] + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \left[\frac{24}{n^5} - \frac {24}{n^52^n} - \frac{24 \ln 2}{n^42^n}-\frac{12\ln^2 2}{n^3 2^n}-\frac{4\ln^3 2}{n^2 2^n} - \frac{\ln^4 2}{n2^n}\right] + \frac {\ln^5 2}{10}\\
=&12\zeta(5) - 12\text{Li}_5(1/2) - 12\ln 2 \text{Li}_4(1/2) -6\ln^2 2 \text{Li}_3(1/2) -2\ln^3 2\text{Li}_2(1/2)-\frac {2}{5}\ln^5 2\\
=&\boxed{-12\Big(\text{Li}_5(1/2) + \ln 2\text{Li}_4(1/2)-\zeta(5)\Big)-{21 \over 4}\zeta(3)\ln^2 2 +{1\over 3} \pi^2 \ln^3 2-{2 \over 5} \ln^5 2}
\end{align*} where the well-known values
\begin{align*}\text{Li}_2(1/2) = {\pi^2 \over 12}-{\ln^2 2\over 2} , \qquad \text{Li}_3(1/2) ={7\zeta(3) \over 8} -{\pi^2 \ln 2\over 12} + {\ln^3 2 \over 6}
\end{align*} are used.
Actually, $I_2$ was already evaluated by the OP here using the algebraic identity $$b^4 + 6a^2b^2 = \frac {(a-b)^4} 2+\frac{(a+b)^4}{2} -a^4.$$
It holds that
$$
\boxed{I_2 = \frac {21}{8} \zeta(5).}
$$
In fact, the value of $I_3$ can also be found in the previous answer of @Przemo's. For $I_3$, one can use the algebraic relation $3(a^3b + ab^3) =\frac 3 8 \left[ (a+b)^4 - (a-b)^4\right]$.
This gives
\begin{align*}
I_3=& \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4(1-x^2)}{x} dx}_{x^2 = y} - \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x} dx}_{\frac{1-x}{1+x} = y}\\
=&\frac 3 {16}\underbrace{\int_0^1 \frac{\ln^4(1-y)}{y} dy }_{1-y\mapsto y}- \frac 3 4 \int_0^1 \frac{\ln^4 y}{1-y^2} dy\\
=&\frac 3 {16}\int_0^1 \frac{\ln^4 y}{1-y} dy - \frac 3 4 \sum_{n=0}^\infty \int_0^1 y^{2n} \ln^4 y \ dy\\
=&\frac 3 {16}\sum_{n=1}^\infty \int_0^1 y^{n-1}\ln^4 y \ dy - \frac 3 4 \sum_{n=0}^\infty \frac {24}{(2n+1)^5}\\
=&\frac 3 {16}\sum_{n=1}^\infty \frac{24}{n^5} - 18 \sum_{n=0}^\infty \frac {1}{(2n+1)^5}\\
=&\frac {9}{2} \zeta(5)- 18\cdot \frac {31}{32}\zeta(5)\\
=&\boxed{-\frac{207}{16}\zeta(5)}
\end{align*} as can be found in @Przemo's answer.
For $I_4$, make substitution $ \frac{1-x}{1+x}\mapsto x$ to get
\begin{align*} I_4 = &2\int_0^1 \frac{\ln^3 x \ln\left(\frac 2 {1+x}\right)}{1-x^2} dx \\
=&2\ln 2 \int_0^1 \frac{\ln^3 x}{1-x^2} dx - \underbrace{2\int_0^1\frac{\ln^3 x \ln(1+x)}{1-x^2} dx }_{=:J}\\
=& 2\ln 2\sum_{n=0}^\infty \int_0^1 x^{2n} \ln^3 x\ dx - J\\
=& - 12\ln 2 \underbrace{\sum_{n=0}^\infty \frac 1 {(2n+1)^4}}_{\frac{15}{16}\zeta(4) = \frac{\pi^4}{96}} - J \\
=& -\frac{\pi^4 \ln 2}{8} - J.
\end{align*}
\begin{align*}
J = &\int_0^1\frac{2\ln^3 x \ln(1+x)}{1-x^2} dx \\
=& \underbrace{\int_0^1 \frac{\ln^3 x \ln(1+x)}{1+x}dx}_{=:A} + \int_0^1 \frac{\ln^3 x \ln(1+x)}{1-x}dx\\
=& A + \int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + \int_0^1 \frac{(1+x)\ln^3 x \ln(1-x^2)}{1-x^2}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + \underbrace{\int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x^2}dx }_{=:B}+\underbrace{\int_0^1 \frac{x\ln^3 x \ln(1-x^2)}{1-x^2}dx}_{x^2 \mapsto x}-\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + B - \underbrace{\frac {15}{16} \int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx}_{=:C}\\
=&A + B - C.
\end{align*}
For $A$, we can use the McLaurin series of
$$
\frac{\ln (1+x)}{1+x} = \sum_{n=0}^\infty (-1)^{n-1}H_n x^n
$$ ($H_0= 0$) to get
\begin{align*}
A = & \sum_{n=0}^\infty (-1)^{n-1}H_n \int_0^1 x^n\ln^3 x \ dx \\
=&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_n}{(n+1)^4}\\
=&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_{n+1}}{(n+1)^4} - 6\sum_{n=0}^\infty \frac{(-1)^{n}}{(n+1)^5}\\
=&6 \sum_{n=1}^\infty \frac{(-1)^{n-1}H_{n}}{n^4} - 6\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^5}\\
=& 6\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-6\cdot \frac{15}{16}\zeta(5)\\
=& \frac{87}{16}\zeta(5) - \frac{\pi^2 \zeta(3)}{2}.
\end{align*}
Here, the known value of $ \sum_{n=1}^\infty (-1)^{n-1}{H_n \over n^4}$ is used.
For $B$, make substitution $u = x^2$ to get
\begin{align*}
B =& \frac 1 {16} \int_0^1 \frac{\ln^3 u \ln(1-u)}{\sqrt u (1-u)} du \\
=& \frac 1 {16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=\frac 1 2, y = 0^+}
\end{align*} where $\text{B}(\cdot,\cdot)$ is Euler's Beta function. We can use the fact that
\begin{align*}
\lim_{y\to 0^+}\frac{\partial^2}{\partial x\partial y} \text{B}(x,y) = -\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]
\end{align*} to get
\begin{align*}
B =& \frac 1 {16}\frac{d^2}{dx^2}\left[-\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]\right]_{x=\frac 1 2}\\
=&\frac 1 {16} \left[-\frac 1 2 \psi''''(1/2) + \psi'''(1/2)\big[\psi(1/2) + \gamma\big] + 3\psi'(1/2)\psi''(1/2)\right]\\
=& \frac 1 {16}\left[-21\pi^2 \zeta(3) + 372\zeta(5) - 2\pi^4 \ln 2\right]
\end{align*} which can be evaluated using the series representations of polygamma functions $$\psi(x) +\gamma = - \frac 1 x +\sum_{n=1}^\infty \frac 1 n - \frac 1 { n+x},\\
\psi'(x) = \sum_{n=0}^\infty \frac 1 {(n+x)^2}$$ and the derived fact that $\psi(\tfrac 1 2 )+\gamma = -2\ln 2$ and $\psi^{(k)}(\tfrac 1 2)=(-1)^{k+1}k!(2^{k+1}-1)\zeta(k+1)$ for $k\ge 1$.
For $C$, we can use the same method as used in the evaluation of $B$. It holds that
\begin{align*}
C =& \frac {15}{16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=1, y = 0^+}\\
=&\frac {15} {16}\left[-\frac 1 2 \psi''''(1) + \psi'''(1)\big[\psi(1) + \gamma\big] + 3\psi'(1)\psi''(1)\right]\\
=&\frac{15}{16}\left[12\zeta(5) -6\zeta(2)\zeta(3)\right]\\
=&\frac {45}{4}\zeta(5) -\frac {15\pi^2 \zeta(3)}{16}
\end{align*} where $\psi(1) +\gamma = 0$, $\psi'(1) = \zeta(2)$, $\psi''(1) = -2\zeta(3)$ and $\psi''''(1) = -24\zeta(5)$ are used.
Combining $A,B,C$, we have that $$J =A+B-C= \frac{279}{16}\zeta(5) -\frac{7\pi^2\zeta(3)}{8} - \frac{\pi^4 \ln 2}{8}$$ and
$$
\boxed{I_4 = -\frac{\pi^4 \ln 2}{8} - J = -\frac{279}{16}\zeta(5)+\frac{7\pi^2\zeta(3)}{8}}
$$
Finally, these evaluate $\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =\frac 1 2\big[-I_1-I_2+I_3-I_4\big]$ as follows.
\begin{align*}
\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =&\ 6\text{Li}_5(1/2) + 6\ln 2\ \text{Li}_4(1/2)-\frac{81}{16}\zeta(5)-{7\pi^2 \over 16}\zeta(3)\\
&+\frac{21\ln^2 2}{8}\zeta(3)- \frac{1}{6}\pi^2\ln^3 2+\frac{1}{5}\ln^5 2.
\end{align*}
Using the identity given in the OP, we get the desired integral $I$
\begin{align*}
\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\
&-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}.
\end{align*}
Best Answer
We shall use $A_n=\sum _{k=1}^n \frac{(-1)^{k+1}}{k}$ (and for completeness $A_0 = 0$) for the alternating harmonic sum.
The sum in question is
$$s = \sum_{k\ge 0} (-1)^k (A_k-\log(2))^2\\= \sum_{k\ge 0} \left((A_{2k}-\log(2))^2-(A_{2k+1}-\log(2))^2\right)$$
The summand can be expanded to
$$A_{2 k}^2-A_{2 k+1}^2-2 \log (2) A_{2 k}+2 \log (2) A_{2 k+1} \\ = -\frac{2 (A_{2 k}-\log (2))}{2 k+1}-\frac{1}{(2 k+1)^2}\tag{1}$$
Here we have used that for integer $k$
$$A_{k+1}-A_{k}=\frac{(-1)^k}{k+1}\tag{2}$$
So we have
$$s = s_1 + s_2$$
with
$$s_1 =2 \sum_{k\ge 0} \frac{ \log(2) - A_{2 k}}{2 k+1}\tag{3}$$
and
$$s_2 = -\sum_{k\ge0} \frac{1}{(2 k+1)^2}= -\frac{\pi^2}{8}\tag{4}$$
which we have summed up imediately.
In the first one we use that for integer $k$
$$A_{2 k}-\log(2) = \int_0^1 \frac{1-x^{2 k}}{x+1} \, dx-\int_0^1 \frac{1}{x+1} \, dx\\ = -\int_0^1 \frac{x^{2 k}}{x+1} \, dx\tag{5}$$
so that the sum under the $x$-integral becomes
$$ \sum_{k\ge0} \frac{2 x^{2 k} }{2 k+1}=\frac{2 \operatorname{arctanh}(x)}{x}=\frac{1}{x} \log(\frac{1+x}{1-x})\tag{6}$$
Hence the first sum is given by the integral
$$s_1 = \int_0^1 \frac{1}{x(1+x)} \log(\frac{1+x}{1-x})=\frac{\pi^2}{6}\tag{7}$$
Finally we get
$$s = s_1 + s_2 =\frac{\pi^2}{6} -\frac{\pi^2}{8}= \frac{\pi^2}{24}$$
Q.E.D.
Discussion
1) Integral representations of the alternating and odd harmonic numbers
Inserting $\frac{1}{k}=\int_0^1 x^{k-1}\,dx$ into the definition of $A_n$ and doing the sum under the integral we obtain
$$A_n=\int_0^1 \frac{1-(-1)^n x^n}{x+1}\,dx\tag{8a}$$
for $n=2k$ this simplifies to
$$A_{2k}=\int_0^1 \frac{1- x^{2k}}{x+1}\,dx\tag{8b}$$
as used in $(5)$.
Similarly we have for the odd harmonic number the representation
$$O_n = \int_0^1 \frac{x^{2 n}-1}{x^2-1} \, dx\tag{9}$$
2) The introduction of an integral in $(4)$ is a deviation from the requirement of the OP to use only series operations.
Ideally we would like to see $s_1= \sum_{k\ge 1} \frac{1}{k^2}$
Hence my "solution" is incomplete. I'll try to improve it.
EDIT
I'm almost there.
I find by series manipulations (changing the order of summation) that
$$s_1 = \sum_{n\ge1} \frac{O_{n}}{(n+1)(2n-1)}\tag{10}$$
Where the odd harmonic sum is defined as
$$O_n=\sum_{m=1}^n \frac{1}{2m-1}\tag{11}$$
It is easy to show that
$$O_n = H_{2n} -\frac{1}{2} H_{n}\tag{12}$$
so that
$$s_1 = \sum_{n\ge1} \frac{H_{2n} -\frac{1}{2} H_{n}}{(n+1)(2n-1)}\tag{13}$$
But we have (https://en.wikipedia.org/wiki/Harmonic_number)
$$H_{2z}=\log(2) + \frac{1}{2}(H_z+H_{z-\frac{1}{2}})\tag{14}$$
so that
$$s_1 = \sum_{n\ge1} \frac{\log(2) -\frac{1}{2} H_{n-\frac{1}{2}}}{(n+1)(2n-1)}\tag{15}$$
The first sum yields $2 \log(2)^2$. The second sum
$$s_{1b} =- \frac{1}{2}\sum_{n\ge1} \frac{ H_{n-\frac{1}{2}}}{(n+1)(2n-1)}= ?\tag{16}$$
Still has to be calculated.
Summarizing: the numerical result of $s_1$ agrees with $\pi^2/6$,