How to find $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$ using real methods?

How to calculate

$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$$

and

$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$$

by means of real methods?

This question was suggested by Cornel the author of the book, Almost Impossible Integrals, Sums and Series.

The way I would approach the problem is to use the series property:

$$\sum_{n=1}^\infty (-1)^n f(2n)=\Re \sum_{n=1}^\infty i^n f(n),$$ namely

$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}=8\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}=8\Re\sum_{n=1}^\infty\frac{i^n H_n}{n^3}$$

then we use the well-known generating function $\sum_{n=1}^\infty\frac{x^nH_n}{n^3}$. But this method is really tedious as we will need to use $\Re\{\text{Li}_2(1+i), \text{Li}_3(1+i),\text{Li}_4(1+i),\text{Li}_4(\frac{1+i}{2})\}.$

By the way, I have not seen a rigorous proof of the following equality:

$$\operatorname{Re} \operatorname{Li}_4 (1 + i)= -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97}{9216} \pi^4 + \frac{\pi^2}{48} \ln^2 2 - \frac{5}{384} \ln^4 2\tag1$$

So solving this sum in a different way would be considered a new rigorous proof of $(1)$.

For the second series, I would follow the same approach.

Any idea by real methods? Thanks