nice two related sums $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}$

Evaluating the first sum: we know the fact that $\int_0^1x^{n-1}\ln(1-x)\ dx=-\frac{H_n}{n}$

differentiate both sides w.r.t $n$, we get $\int_0^1 x^{n-1}\ln x\ln(1-x)\ dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}$

now multiply both sides by $\frac{(-1)^n}{n}$ then take the sum, we get \begin{align} \sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n^3}+\frac{H_n^{(2)}}{n^2}-\frac{\zeta(2)}{n^2}\right)&=\int_0^1\frac{\ln x\ln(1-x)}{x}\sum_{n=1}^\infty\frac{(-x)^n}{n}\ dx\\ &=-\int_0^1\frac{\ln x\ln(1-x)\ln(1+x)}{x}\ dx \end{align} I proved here $$\int_0^1\frac{\ln x\ln(1-x)\ln(1+x)}{x}\ dx=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)$$ giving us $$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^2}=-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}+\zeta(2)\sum_{n=1}^\infty\frac{(-1)^n}{n^2}-\frac18\zeta(4)$$

Using the well known results

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42\tag1$$

$$\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)\tag2$$

$$ \sum_{n=1}^\infty\frac{(-1)^n}{n^2}=\operatorname{Li}_2(-1)=-\frac12\zeta(2)$$

Thus $$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42$$

Note that $(1)$ is calculated here and $(2)$ can be found using Euler Identity.

Evaluating the second sum:

using the second derivative of beta function, we have $\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2}{n}+\frac{H_n^{(2)}}{n}$

multiply both sides by $\frac{(-1)^n}{n}$ then take the sum, we get \begin{align} \sum_{n=1}^\infty(-1)^n\left(\frac{H_n^2}{n^2}+\frac{H_n^{(2)}}{n^2}\right)&=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{(-1)^n}{n}\ dx\\ &=-\int_0^1\frac{\ln^2(1-x)\ln(1+x)}{x}\ dx\\ &=-\int_0^1\frac{\ln^2x\ln(2-x)}{1-x}\ dx\\ &=-\ln2\int_0^1\frac{\ln^2x}{1-x}\ dx-\int_0^1\frac{\ln^2x\ln(1-x/2)}{1-x}\ dx\\ &=-2\ln2\zeta(3)+\sum_{n=1}^\infty\frac{1}{n2^n}\int_0^1\frac{x^n\ln^2x}{1-x}\ dx\\ &=-2\ln2\zeta(3)+\sum_{n=1}^\infty\frac{1}{n2^n}\left(2\zeta(3)-2H_n^{(3)}\right)\\ &=-2\ln2\zeta(3)+2\ln2\zeta(3)-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}\\ &=-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n} \end{align} rearanging the terms, we get $$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}=-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^2}-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}$$ using the generating function $\sum_{n=1}^\infty\frac{x^nH_n^{(3)}}{n}=\operatorname{Li_4}(x)-\ln(1-x)\operatorname{Li_3}(x)-\frac12\operatorname{Li_2}^2(x)$, we can find: $$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$ Finally, plugging these two sums, we get $$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$

A different way can be found here.