The closed-form of $\sum_{n=0}^{\infty}\frac{(-1)^n H^{(2)}_{n}}{(2n+1)^2} $
Solution by Cornel Valean.
\begin{align} S&=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^2}\\ &=\sum_{n=1}^\infty(-1)^{n-1}H_n^{(2)}\int_0^1-x^{2n}\ln x\ dx\\ &=\int_0^1\ln x\sum_{n=1}^\infty(-x^2)^nH_n^{(2)}\ dx\\ &=\int_0^1\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\ &=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx-\underbrace{\int_1^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx}_{x\mapsto 1/x}\\ &=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\int_0^1\frac{\ln x\operatorname{Li}_2(-1/x^2)}{1+x^2}\ dx\\ 2S&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\int_0^1\frac{\ln x[\color{red}{\operatorname{Li}_2(-x^2)+\operatorname{Li}_2(-1/x^2)}]}{1+x^2}\ dx\\ S&=\frac12\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\frac12\int_0^1\frac{\ln x[\color{red}{-2\ln^2x-\zeta(2)}]}{1+x^2}\ dx\\ &=\frac12\underbrace{\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx}_{I}-\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}\ dx}_{-6\beta(4)}-\frac12\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x^2}\ dx}_{-G}\\ &=\frac12I+6\beta(4)+\frac12G\zeta(2)\tag1 \end{align}
\begin{align} I&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\ &=\int_0^\infty\frac{\ln x}{1+x^2}\left(\int_0^1\frac{x^2\ln y}{1+yx^2}\ dy\right)\ dx\\ &=\int_0^1\ln y\left(\int_0^\infty\frac{x^2\ln x}{(1+x^2)(1+yx^2)}\ dx\right)\ dy\\ &=\int_0^1\frac{\ln y}{1-y}\left(\int_0^\infty\frac{\ln x}{1+yx^2}\ dx-\underbrace{\int_0^\infty\frac{\ln x}{1+x^2}\ dx}_{0}\right)\ dy\\ &=\int_0^1\frac{\ln y}{1-y}\left(-\frac{\pi}{4}.\frac{\ln y}{\sqrt{y}}\right)\ dy,\quad \sqrt{y}=x\\ &=-2\pi\int_0^1\frac{\ln^2x}{1-x^2}\ dx=-2\pi\left(-\frac74\zeta(3)\right)=\boxed{-\frac72\pi\zeta(3)}\tag2 \end{align}
Plug (2) and (1) we get
$$S=-\frac74\pi\zeta(3)+6\beta(4)+\frac12G\zeta(2)$$
where $\beta(4)=\frac{\psi_3(1/4)}{768}-\frac{15}{16}\zeta(4)$
Addendum:
Another approach is by applying integration by parts to the integral $S$ then we change the limits from $(0,1)$ to $(0,\infty)$ as we did in our solution above.