How can we show that : $\sum\limits_{n=1}^{\infty}{H_{n,2}\over n^2}={7\over4}\cdot\zeta(4)$

One may use multizeta values, $$ \zeta(a,b)=\sum_{n_1> n_2>0}\frac{1}{n_1^{a} n_2^{b}}=\sum_{n=1}^\infty \frac{H_{n,b}}{(n+1)^a} $$ then we have Euler's reflection formula $$ \zeta(a,b)+\zeta(b,a)=\zeta(a)\zeta(b)-\zeta(a+b) $$ here, $a=b=2$ gives $$ \zeta(2,2)=\sum_{n=1}^\infty \frac{H_{n,2}}{(n+1)^2}=\frac{(\zeta(2))^2-\zeta(4)}2=\frac{\frac{\pi ^4}{36}-\frac{\pi ^4}{90}}2=\frac34\cdot \zeta(4) $$ as wanted.

Going back to Euler's solution to the Basel problem, we have in fact (see here, pp. 6-7),

$$ \sum_{n=0}(-1)^n\zeta(\underbrace{2,2,\cdots,2,2}_{n})\cdot t^{2n}=\prod_{n=1}^\infty\left(1-\frac{t^2}{n^2} \right)=\frac{\sin (\pi t)}{\pi t}. $$

yielding $$ \zeta(\underbrace{2,2,\cdots,2,2}_{n})=\frac{\pi^{2n}}{(2n+1)!},\quad n\ge1. $$

That is fairly easy with a simmetry argument given by the following Lemma: $$ \sum_{n=1}^{N}f(n)\sum_{m=1}^{n} f(m) = \!\!\!\sum_{1\leq m\leq n\leq N}f(n)\,f(m)=\frac{1}{2}\left(\left(\sum_{n=1}^{N}f(n)\right)^2+\sum_{n=1}^{N}f(n)^2\right) \tag{1}$$ If we consider $f(n)=\frac{1}{n^2}$ and take the limit as $N\to +\infty$, we instantly get: $$ \sum_{n\geq 1}\frac{H_n^{(2)}}{n^2} = \frac{\zeta(2)^2+\zeta(4)}{2} = \color{red}{\frac{7}{4}\,\zeta(4)}\tag{2}$$ as wanted.

Yes, it is correct. More generally, for $k>1$ and for $N>0$ $$H_{N,k}^2=2\sum_{n=1}^{N}{H_{n-1,k}\over n^k}+H_{N,2k}$$ and then by taking the limit we get $$\zeta(k)^2=2\sum_{n=1}^{\infty}{H_{n-1,k}\over n^k}+\zeta(2k).$$ Hence $$\sum_{n=1}^{\infty}{H_{n,k}\over n^k}= \sum_{n=1}^{\infty}{H_{n-1,k}\over n^k}+\zeta(2k)=\frac{\zeta(k)^2+\zeta(2k)}{2}.$$