A partial answer was provided in response to my MSE question,
"ArcTan(2) a rational multiple of $\pi$?"

There Thomas Andrews showed that $\arctan(x)$ is not a rational multiple of $\pi$ for any
$x$ rational, except for $-1,0,1$.
More specifically:

$\arctan(x)$ is a rational multiple of $\pi$ if and only if the complex number $1+xi$ has the property that $(1+xi)^n$ is a real number for some positive integer $n$.
This is not possible if $x$ is a rational,
$|x|\neq 1$, because $(q+pi)^n$ cannot be real for any $n$ if $(q,p)=1$ and $|qp|> 1$. So $\arctan(\frac{p}q)$ cannot be a rational multiple of $\pi$.