Does $\sum_{n=2}^\infty \frac{1}{(\ln(n))^2}$ converge or diverge?

Observe that $x > (\ln x)^2 $ and so

$$ \frac{1}{(\ln x)^2} > \frac{1}{x} $$

for $x>1$. Applying this inequality to your case, you can see that the given numerical series is divergent

One can prove that any sum of the form $$\sum_{k=2}^\infty\frac1{\ln^n{(k)}}$$ diverges where $n\in\mathbb{R}$. Firstly, it can be proven by induction and L'Hôpital's rule that $$\lim_{x\to\infty}\frac{x^n}{e^x}=0$$ for all $n\in\mathbb{R}$. Then by using this fact we must have $$x^n\lt e^x$$ for all $x\gt x_0$ as $x^n$ is increasing and using the definition of a limit. Taking $x=\ln{(t)}$ for some $t\in\mathbb{R}$ we get $$\ln^n{(t)}\lt t$$ for all $t\gt t_0$ where $t_0=e^{x_0}$. Thus for all $k\ge \lceil t_0\rceil$ we have that $$\ln^n{(k)}\lt k$$ and hence $$\frac1{\ln^n{(k)}}\gt \frac1k$$ So the sum $$\sum_{k=\lceil t_0\rceil}^\infty \frac1{\ln^n{(k)}}\gt\sum_{k=\lceil t_0\rceil}^\infty\frac1k\to\infty$$ Hence $$\sum_{k=2}^\infty\frac1{\ln^n{(k)}}\gt\sum_{k=\lceil t_0\rceil}^\infty \frac1{\ln^n{(k)}}\to\infty$$