Perhaps it is in the extraction of the values you are having trouble with. If so, here is one rather (tedious and laborious) way to do it.
From Eq. (590) in the link, one has
$$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \operatorname{Li}_2 (x) + \frac{1}{2} \ln^2 (1 - x).$$
Setting $x = e^{i\pi /3}$ we see that
\begin{align}
\sum_{n = 1}^\infty \frac{H_n}{n} \cos \left (\frac{n \pi}{3} \right ) &= \operatorname{Re} \sum_{n = 1}^\infty \frac{H_n}{n} e^{i \pi n/3}\\ &= \operatorname{Re} \left [\operatorname{Li}_2 \left (e^{i\pi/3} \right ) + \frac{1}{2} \ln^2 \left (1 - e^{i \pi/3} \right ) \right ].
\end{align}
The log term can be disposed of immediately. Here
$$\ln (1 - e^{i \pi/3}) = -\frac{i \pi}{3}.$$
For the dilogarithm term, $\operatorname{Re} \operatorname{Li}_2 (e^{i \theta})$ is an elementary function (for a reason why, see here) and means a value for this term can be found. What I present below is one way to find it (and there are no doubt other, simpler ways).
Starting with the series representation for the dilogarithm function, we have
\begin{align}
\operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] &= \operatorname{Re} \sum_{n = 1}^\infty \frac{e^{in \pi/3}}{n^2}\\
&= \sum_{n = 1}^\infty \frac{\cos \left (\frac{n \pi}{3} \right )}{n^2}\\
&= \sum_{\substack{n = 1\\n \in 6,12,\ldots}}^\infty \frac{1}{n^2} - \sum_{\substack{n = 1\\n \in 3,9,\ldots}}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{\substack{n = 1\\n \in 1,7,\ldots}}^\infty \frac{1}{n^2}\\
& \quad - \frac{1}{2} \sum_{\substack{n = 1\\n \in 2,8,\ldots}}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{\substack{n = 1\\n \in 5,11,\ldots}}^\infty \frac{1}{n^2} - \frac{1}{2} \sum_{\substack{n = 1\\n \in 4,10,\ldots}}^\infty \frac{1}{n^2}.
\end{align}
Terms in the series can be rearranged as the series absolutely converges. Shifting the indices as follows: $n \mapsto 6n$, $n \mapsto 6n + 3$, $n \mapsto 6n +1$, $n \mapsto 6n + 2$, $n \mapsto 6n + 5$, $n \mapsto 6n + 4$ leads to
\begin{align}
\operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] &= \frac{1}{36} \sum_{n =1}^\infty \frac{1}{n^2} - \frac{1}{36} \sum_{n = 0}^\infty \frac{1}{(n + 1/2)^2} + \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 1/6)^2}\\
& \quad - \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 1/3)^2} + \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 5/6)^2} - \frac{1}{72} \sum_{n = 0}^\infty \frac{1}{(n + 2/3)^2}\\
&= \frac{1}{72} \left [2 \cdot \zeta (2) - 2 \psi^{(1)} \left (\frac{1}{2} \right ) + \psi^{(1)} \left (\frac{5}{6} \right ) + \psi^{(1)} \left (\frac{1}{6} \right ) - \psi^{(1)} \left (\frac{2}{3} \right ) - \psi^{(1)} \left (\frac{1}{3} \right ) \right ],
\end{align}
where $\psi^{(1)} (z)$ is the polygamma function (trigamma function).
Using the known values of $\zeta (2) = \pi^2/6$, $\psi^{(1)} (1/2) = \pi^2/2$, and making use of the reflection relation for the trigamma function of
$$\psi^{(1)} (1 - z) + \psi^{(1)} (z) = \pi^2 \csc^2 (\pi z),$$
we see that
$$\operatorname{Re} \left [\operatorname{Li}_2 (e^{i \pi/3}) \right ] = \frac{1}{72} \left [2 \cdot \frac{\pi^2}{6} - 2 \cdot \frac{\pi^2}{2} + 4 \pi^2 - \frac{4 \pi^2}{3} \right ] = \frac{\pi^2}{36}.$$
Thus
$$\sum_{n = 1}^\infty \frac{H_n}{n} \cos \left (\frac{n \pi}{3} \right ) = \frac{\pi^2}{36} - \frac{\pi^2}{18} = -\frac{\pi^2}{36},$$
as desired.
A second solution in large steps by Cornel Ioan Valean
Let's start with the following useful identity which is easily derived by using recurrence relations and simple rearrangements, manipulations with sums, that is
Let $n$ be a non-negative integer number. Then, we have
$$\int_0^1 x^{2n}\frac{\log(1+x)}{1+x}\textrm{d}x$$
$$=\frac{1}{2}H_{2n}^2-2\log(2) H_{2n}+\frac{1}{2}H_{2n}^{(2)}-\frac{1}{4}H_n^2-\frac{1}{4}H_n^{(2)}+\log (2)H_n+\frac{1}{2} \log ^2(2)-\sum_{k=1}^{n-1}\frac{H_k}{2 k+1},$$
where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$.
By multiplying both sides of the identity above by $1/n^3$ and considering the summation from $n=1$ to $\infty$, we get
$$\sum_{n=1}^{\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{H_{k}}{2 k+1}=\sum_{k=1}^{\infty} \sum_{n=k+1}^{\infty}\frac{1}{n^3}\frac{H_{k}}{2 k+1}=\underbrace{\sum_{k=1}^{\infty}\frac{H_{k}}{2 k+1}\left(\zeta(3)-H_k^{(3)}\right)}_{\text{The desired series}}$$
$$=\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^2}{n^3}-2\log(2) \sum_{n=1}^{\infty}\frac{H_{2n}}{n^3}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^{(2)}}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty} \frac{H_n^{(2)}}{n^3}$$
$$+\log (2)\sum_{n=1}^{\infty} \frac{H_n}{n^3}+\frac{1}{2}\log ^2(2)\sum_{n=1}^{\infty}\frac{1}{n^3}-\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x,$$
where we see all the series in the right-hand side are easily reducible to known series which may also be found in the book (Almost) Impossible Integrals, Sums, and Series.
On the other hand, with simple integration by parts, we obtain
$$\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x$$
$$=\frac{1}{2}\log^2(2)\zeta(3)-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(x)}{x}\textrm{d}x-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(-x)}{x}\textrm{d}x,$$
where the last integrals may be found calculated in the paper The calculation of a harmonic series with a weight $5$ structure, involving the product of harmonic numbers, $H_n H_{2n}^{(2)}$.
A note: The sister of the result above (easy to obtain by recurrence relations and very useful),
$$\int_0^1 x^{2n-1} \frac{\log(1+x)}{1+x}\textrm{d}x$$
$$=2\log(2) H_{2n}-\log(2)H_n+\frac{1}{4}H_n^2+\frac{1}{4}H_n^{(2)}-\frac{1}{2}H_{2n}^2-\frac{1}{2} H_{2n}^{(2)}+\frac{H_{2n}}{2n}-\frac{H_n}{2n}
$$
$$
-\frac{1}{2}\log^2(2)+\sum_{k=1}^{n-1}\frac{H_k}{2 k+1}.
$$
Best Answer
Beside telescoping, all you need is to reverse the order of the double sum: $$\sum_{n\ge 1}\frac{\sum_{k=1}^n k^{-1}}{n(n+1)}=\sum_{k\ge 1}k^{-1}\sum_{n\ge k}\frac{1}{n(n+1)}=\sum_{k\ge 1}k^{-2}=\frac{\pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $\frac{k^{-1}}{n(n+1)}$ with $n,\,k$ integers in the set $\{(n,\,k)|n,\,k\ge 1,\,k\le n\}$. But I could equivalently describe this set as $\{(n,\,k)|k\ge 1,\,n\ge k\}$.