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:

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

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.