Solve:
$2 \cdot \frac{2 \pi}{3}+\varphi=k \pi+\frac{\pi}{2} \Rightarrow \varphi=k \pi-\frac{5 \pi}{6} \Rightarrow \varphi=\frac{\pi}{6}$, thus $f(0)=\frac{M}{2}$, option $A$ is incorrect; also, $M$ can be positive or negative, so options $B, D$ are incorrect; and $2 \cdot \frac{5 \pi}{12}+\frac{\pi}{6}=\pi$ satisfies the condition, so the answer is $C$.
3. Let the function $f(x)=M \sin (\omega x+\varphi)\left(M \neq 0, \omega>0,-\frac{\pi}{2}<\varphi<\frac{\pi}{2}\right)$ have a graph that is symmetric about the line $x=\frac{2 \pi}{3}$, and its period is $\pi$, then
A. The graph of $f(x)$ passes through the point $\left(0, \frac{1}{2}\right)$
B. $f(x)$ is a decreasing function on $\left[\frac{\pi}{12}, \frac{2 \pi}{3}\right]$
C. A center of symmetry for $f(x)$ is $\left(\frac{5 \pi}{12}, 0\right)$
D. The maximum value of $f(x)$ is $M$