Is $\arcsin(1/4) / \pi$ irrational?

Yes, $\arcsin(\frac14)/\pi$ is irrational.

Suppose $\arcsin(\frac14)/\pi = m/n$, where $m$ and $n$ are integers.

Then $\sin(n \arcsin(\frac14))=\sin(m \pi)=0$.

We analyze this usng the formulas from Browmich as cited on Mathworld:

$$\frac{\sin(n\arcsin(x))}{n}=x-\frac{(n^2-1^2)x^3}{3!} + \frac{(n^2-1^2)(n^2-3^2)x^5}{5!} + \cdots$$ $$\frac{\sin(n\arcsin(x))}{n \cos(\arcsin(x))}=x-\frac{(n^2-2^2)x^3}{3!} + \frac{(n^2-2^2)(n^2-4^2)x^5}{5!} + \cdots$$ for $n$ odd or even respectively.

So the right-hand sides must be 0 for $x=\frac14$.

However, when we multiply the terms on the right-hand sides by $2^nn$ (if $n$ is odd) or $2^{n-2}n$ (if $n$ is even), we find that all the terms are integers, except that the last non-zero term is $\pm\frac12$.

So the right-hand side can't be 0, the left-hand side can't be 0, and $\arcsin(\frac14)/\pi$ must be irrational.

This is a partial case of the classical result.

https://en.wikipedia.org/wiki/Niven%27s_theorem

Let $\theta= \arcsin(1/4)$. Assume $\theta$ is a rational multiple of $\pi$.

Then, there exists some $n$ such that $\sin(n\theta)=0$. This gives $\cos(n \theta)= \pm 1$.

Set $z=\cos(\theta)+i \sin(\theta)$, then $z^n= \pm 1$ and $\frac{1}{z^n}=\pm 1$.

This gives that $z$ and $\frac{1}{z}$ are algebraic integers, and hence so is $$2 i\sin(\theta)=z- \frac{1}{z}$$

Therefore, $\frac{i}{2}$ is an algebraic integer, which is a contradiction.