Denoting $H_0=0$, we have $$\sum_{n=1}^\infty \frac{H_n}{n(n+1)2^n}=\sum_{n=1}^\infty \left(\frac1n-\frac1{n+1}\right)\frac{H_n}{2^n}=\sum_{n=1}^\infty \frac{1}n\left(\frac{H_n}{2^n}-\frac{H_{n-1}}{2^{n-1}}\right)\\=\sum_{n=1}^\infty\frac1{n^22^n}-\sum_{n=1}^\infty\frac{H_n}{(n+1)2^{n+1}}.$$
It is well known that $\sum_{n=1}^\infty\frac1{n^22^n}=\frac{\pi^2}{12}-\frac{\log^2 2}2$ (see the value of ${\rm Li}_2(1/2)$ here). Thus, it remains to show that $$\sum_{n=1}^\infty\frac{H_n}{(n+1)2^n}=\log^2 2.$$
For this, we take the square of the series $$
\log 2=-\log\left(1-\frac12\right)=\frac12+\frac1{2\cdot 2^2}+\frac1{3\cdot 2^3}+\ldots$$
to get $$\log^2 2=\sum_{a,b=1}^\infty \frac1{ab2^{a+b}}=\sum
_{a,b=1}^\infty \frac1{(a+b)2^{a+b}}\left(\frac1a+\frac1b\right)=\sum_{n=2}^\infty\frac1{n2^n}2H_{n-1}=\sum_{n=1}^\infty\frac{H_n}{(n+1)2^{n}}.$$