Note that for $k=1,2,3$
$$\cos\left(\frac{2\pi k}{3}\right)= -\frac1 2, -\frac1 2,1$$
and it is periodic therefore
$$S_N=\sum_{k=1}^{3N}\frac{\cos(\frac{2\pi k}{3})}{k^2}=\sum_{k=0}^{N-1}\frac{\cos(\frac{2\pi (3k+1)}{3})}{(3k+1)^2}+\sum_{k=0}^{N-1}\frac{\cos(\frac{2\pi (3k+2)}{3})}{(3k+2)^2}+\sum_{k=0}^{N-1}\frac{\cos(\frac{2\pi (3k+1)}{3})}{(3k+3)^2}=$$
$$=-\frac1 2\sum_{k=0}^{N-1}\frac{1}{(3k+1)^2}-\frac 12\sum_{k=0}^{N-1}\frac{1}{(3k+2)^2}+\sum_{k=0}^{N-1}\frac{1}{(3k+3)^2}$$
then we can use the following idea to obtain
$$=-\frac1 2\sum_{k=0}^{N-1}\frac{1}{(3k+1)^2}-\frac 12\sum_{k=0}^{N-1}\frac{1}{(3k+2)^2}-\frac12\sum_{k=0}^{N-1}\frac{1}{(3k+3)^2}+\frac{3}{2}\sum_{k=0}^{N-1}\frac{1}{(3k+3)^2}=$$
$$S_N=-\frac1 2\sum_{k=0}^{N-1}\frac{1}{(k+1)^2}+\frac{1}{6}\sum_{k=0}^{N-1}\frac{1}{(k+1)^2}=-\frac{1}{3}\sum_{k=0}^{N-1}\frac{1}{(k+1)^2}\to -\frac{\pi^2}{18}$$
For the other one, in a similar way we obtain
$$S_N=\sum_{k=1}^{3N}\frac{\sin\left(\frac{2\pi k}{3}\right)}{k^2}=\frac{\sqrt 3} 2\sum_{k=0}^{N-1}\frac{1}{(3k+1)^2}-\frac{\sqrt 3} 2\sum_{k=0}^{N-1}\frac{1}{(3k+2)^2}$$
which in the limit gives by Polygamma function
$$S_N\to \frac{\sqrt 3} {18} \left(\psi'\left(\frac13\right)-\psi'\left(\frac23\right)\right)\approx \frac23$$