Why cannot $[0;\pi]$ be range of $\arcsin$?
Because at this range $\sin$ is not injective. It means that there exist $a,b\in[0;\pi],\;a\ne b$, that $\sin(a)=\sin(b)$. This is very inconvenient because $\arcsin$ would be multivalued. For one argument there would exist two values. That's why such range is selected that $\sin$ is injective and thus $\arcsin$ is a function.
Function $\sin(x)$ is periodic. It means that for every value $y$ there exist infinitely many arguments $x$ satisfying $y=\sin(x)$. Then, its inverse $\arcsin$ is multivalued. For every argument it takes infinitely many values. That's why we usually "split" it into branches, so that every branch is a function. It is usually done at points where tangent is vertical. Then we define such branches as $$...,\;\arcsin_{-1}(x),\;\arcsin_0(x),\;\arcsin_1(x),...$$
The branch $\arcsin_0(x)$ is the principal one, equivalent to $\arcsin(x)$.
Reasoning behind $\arccos$ is analogous.
Similar thing goes with $\tan$. It is not injective, so we take such domain (this selected domain of $\tan$ is the range of $\arctan$) that $\tan$ is injective on it. It is $$...,\;(-3\pi/2;-\pi/2),\;(-\pi/2;\pi/2),\;(\pi/2,3\pi/2),\;...$$
Actually we can take domains such as $[0;\pi)$, but then, $\arctan$ would not be continuous.
Why the domain that is closest to $0$ is selected? Because it's convenient.
