# is the $\arcsin$ of a transcendental number, algebraic?

It's know that $\sin a$ is a transcendental $t$ if $a \neq 0$ and algebraic. So

$\sin a = t$

This would imply

$\arcsin(t) = a$

And this question can be done for all inverse trigonometry.

$2^i$ is transcendental by the Gelfond-Schneider theorem, so $\sin(\ln(2)) = \frac{2^{i} - 1/2^{i}}{2i}$ is transcendental; $\ln(2)$ is transcendental by Lindemann's theorem. Thus $x = \sin(\ln(2))$, $\arcsin(x) = \ln(2)$ is an example where $x$ and $\arcsin(x)$ are both transcendental.
No. The set of algebraic numbers is countable, while the set of transcendental numbers is not. Hence, there is at least one (and actually uncountable many!) transcendental numbers $\alpha$ for which $\sin^{-1} \alpha$ is also transcendental. This follows, e.g. from the continuity and monotonicity of the arcsin function.
You are trying to take the converse of the implication, which is not valid. It is true that if $a \neq 0$ is algebraic, then $\sin a$ is transcendental. However, there are transcendental $t$ such that $\arcsin t$ is also transcendental. As there are uncountably many transcendental numbers and only countably many algebraic numbers, most transcendental numbers (less than $1$ in absolute value) have transcendental $\arcsin$. It is hard to prove specific numbers transcendental. I would be "sure" that $\arcsin \frac e{10}$ is transcendental, but I don't have an approach to prove it.