= Solution
Away from integer multiples of $\pi$, the function is a composition of <continuous functions>. At zero, the bound $|g(x)|\leq\sqrt{|x|}$ gives $g(x)\to0=g(0)$ by the <squeeze theorem>. This is <oscillation multiplied by a vanishing amplitude>.
Fix $n\ne0$ and set $t_k=\pi/2+2\pi k$ and $x_k=n\pi+(-1)^n\arcsin(1/t_k)$. Then $x_k\to n\pi$, $\sin x_k=1/t_k$, and
$$
g(x_k)=\sqrt{|x_k|}\longrightarrow\sqrt{|n\pi|}\ne0=g(n\pi).
$$
So $g$ is discontinuous at every nonzero integer multiple of $\pi$. Its \b[set of continuity points] is
$$
\boxed{\mathbb R\setminus\{n\pi:n\in\mathbb Z,\ n\ne0\}.}
$$
On $[0,\pi]$, $g(x)\leq\sqrt{x}<\sqrt\pi$ for $x<\pi$, while $g(\pi)=0$. With $x_k=\pi-\arcsin(1/t_k)$ as above, $g(x_k)=\sqrt{x_k}\to\sqrt\pi$. Consequently \b[the supremum is $\sqrt\pi$ and is not attained]. This illustrates that <supremum can fail to be attained at an oscillatory endpoint>.
On $[\pi,3\pi/2]$, set $x_0=\pi+\arcsin(2/(3\pi))$. It lies strictly inside the interval, and $1/\sin x_0=-3\pi/2$, so $g(x_0)=\sqrt{x_0}>\sqrt\pi$. Choose $0<\delta<x_0-\pi$. On $[\pi,\pi+\delta]$,
$$
g(x)\leq\sqrt{\pi+\delta}<\sqrt{x_0}=g(x_0).
$$
On $[\pi+\delta,3\pi/2]$, $g$ is a <continuous function>, so the <extreme value theorem> gives a maximum there, at least $g(x_0)$. It exceeds every value on the discarded interval. Therefore \b[$g$ does attain its supremum on $[\pi,3\pi/2]$]. This is the <interior-value criterion for an attained maximum>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-1-oscillatory-suprema.png]
{title=Oscillatory function near pi: an unattained left supremum and a right interior value above the endpoint envelope}
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