To compute these two sums, we are going to establish two relations and solve them by elimination.
First Relation:
From here we have
$$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$
Multiply both sides by $\frac{H_n}{n}$ then sum both sides from $n=1$ to $\infty$ to get
\begin{align}
R_1&=\sum_{n=1}^\infty\frac{H_n^4}{n^2}+3\sum_{n=1}^\infty\frac{H_n^2 H_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=-\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{H_n}{n}x^n\ dx\\
&=-\int_0^1\frac{\ln^3(1-x)}{x}\left(\frac12\ln^2(1-x)+\operatorname{Li}_2(x)\right)\ dx\\
&=-\frac12\int_0^1\frac{\ln^5(1-x)}{x}\ dx-\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_2(x)}{x}\ dx\\
&=60\zeta(6)-I
\end{align}
Then $$\boxed{R_1=\sum_{n=1}^\infty\frac{H_n^4}{n^2}+3\sum_{n=1}^\infty\frac{H_n^2 H_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=60\zeta(6)-I}$$.
Second Relation:
From here, we have
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
Divide both sides by $x$ then integrate from $x=0$ to $x=y$, we get
$$-\int_0^y\frac{\ln^3(1-x)}{x(1-x)}\ dx=\sum_{n=1}^\infty \frac{y^n}{n}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
Now multiply both sides by $-\frac{\ln(1-y)}{y}$ then integrate from $y=0$ to $y=1$ and use the fact that $-\int_0^1 y^{n-1}\ln(1-y)\ dy=\frac{H_n}{n}$, we get
\begin{align}
R_2&=\sum_{n=1}^\infty\frac{H_n^4}{n^2}-3\sum_{n=1}^\infty\frac{H_n^2 H_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\int_0^1\int_0^y\frac{\ln^3(1-x)\ln(1-y)}{xy(1-x)}\ dx\ dy\\
&=\int_0^1\frac{\ln^3(1-x)}{x(1-x)}\left(\int_x^1\frac{\ln(1-y)}{y}\ dy\right)\ dx\\
&=\int_0^1\frac{\ln^3(1-x)}{x(1-x)}\left(\operatorname{Li}_2(x)-\zeta(2)\right) dx\\
&=\int_0^1\frac{\ln^3(1-x)}{x}\left(\operatorname{Li}_2(x)-\zeta(2)\right) dx+\underbrace{\int_0^1\frac{\ln^3(1-x)}{1-x}\left(\operatorname{Li}_2(x)-\zeta(2)\right) dx}_{IBP}\\
&=I-\zeta(2)\int_0^1\frac{\ln^3(1-x)}{x}\ dx-\frac14\int_0^1\frac{\ln^5(1-x)}{x}\ dx\\
&=I+6\zeta(2)\zeta(4)+30\zeta(6)\\
&=I+\frac{81}{2}\zeta(6)
\end{align}
Then $$\boxed{R_2=\sum_{n=1}^\infty\frac{H_n^4}{n^2}-3\sum_{n=1}^\infty\frac{H_n^2 H_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=I+\frac{81}{2}\zeta(6)}$$.
Therefore
\begin{align}
\sum_{n=1}^\infty\frac{H_n^4}{n^2}&=\frac{R_1+R_2}{2}=\frac{201}{4}\zeta(6)-2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}\\
&=\frac{201}{4}\zeta(6)-2\left(\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)\right)\\
&=\frac{797}{24}\zeta(3)+3\zeta^2(3)
\end{align}
where the value of $\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}$ is calculated here.
and
\begin{align}
\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}&=\frac{R_1-R_2}{6}=\frac{13}{4}\zeta(6)-\frac13I
\end{align}
and the remaining integral can be approached as follows:
\begin{align}
I&=\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_2(x)}{x}\ dx\overset{1-x\ \mapsto\ x}{=}\int_0^1\frac{\ln^3x\operatorname{Li}_2(1-x)}{1-x}\ dx\\
&=\int_0^1\frac{\ln^3x}{1-x}(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x))\ dx\\
&=\zeta(2)\int_0^1\frac{\ln^3x}{1-x}\ dx-\int_0^1\frac{\ln^4x\ln(1-x)}{1-x}\ dx-\int_0^1\frac{\ln^3x\operatorname{Li}_2(x)}{1-x}\ dx\\
&=-6\zeta(2)\zeta(4)+\sum_{n=1}^\infty H_n\int_0^1x^n\ln^4x\ dx-\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n \ln^3x \ dx\\
&=-\frac{21}2\zeta(6)+24\sum_{n=1}^\infty\frac{H_n}{(n+1)^5}+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^4}\\
&=-\frac{21}2\zeta(6)+24\sum_{n=1}^\infty\frac{H_n}{n^5}-24\zeta(6)+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}-6\zeta(6)\\
\end{align}
Substituting $\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$ and $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}=\zeta^2(3)-\frac13\zeta(6)$ ( proved here), we get
$$I=-\frac12\zeta(6)-6\zeta^2(3)$$
Consequently
$$\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}=\frac{41}{12}\zeta(6)+2\zeta^2(3)$$