integration - Compute $sum_{n=1}^inftyfrac{H_n^{(2)}}{n^7}$ and $sum_{n=1}^inftyfrac{H_n^2}{n^7}$

How to prove that

$$S_1=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}=7\zeta(2)\zeta(7)+2\zeta(3)\zeta(6)+4\zeta(4)\zeta(5)-\frac{35}{2}\zeta(9)\ ?$$
$$S_2=\sum_{n=1}^\infty\frac{H_n^2}{n^7}=-\zeta(2)\zeta(7)-\frac72\zeta(3)\zeta(6)+\frac13\zeta^3(3)-\frac{5}{2}\zeta(4)\zeta(5)+\frac{55}{6}\zeta(9)\ ?$$
where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

I came across these two sums while working on an tough one of wight 9 and I managed to find these two results but I don't think my solution is a good approach as it's pretty long and involves tedious calculations, so I am seeking different methods if possible. I am much into new ideas. All approaches are appreciated though.

By the way, do we have a generalization for $\displaystyle\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^a}$, for odd $a$? Note that there is no closed form for even $a>4$.

Thanks in advance.

Note: You can find these two results on Wolfram Alpha here and here respectively but I modified it a little bit as I like it expressed in terms of $\zeta(a)$ instead of $\pi^a$.