12. Given a sine curve $y=A \sin (\omega x+\varphi)(A>0, \omega>0,0<\varphi<\pi)$ passing through three points in the same period: $A(0,2), B\left(\frac{\overline{5}}{2}, 0\right), C(3,-2)$, the function expression that meets the conditions is $\qquad$ .
12. Given points $A, B, C$ on the curve, we have
$A \sin \varphi=2$,
$$
\begin{array}{l}
A \sin \left(\frac{5}{2} \omega+\varphi\right)=0, \\
A \sin (3 \omega+\varphi)=-2 .
\end{array}
$$

From (2), we get
$$
\frac{5}{2} \omega+\varphi=k_{1} \pi\left(k_{1} \in \mathbf{Z}\right) \text {. }
$$

Since points $A(0,2)$ and $B(3,-2)$ are within the same period, the smallest positive period $T>3$. That is, $T=\frac{2 \pi}{\omega}>3$, from which we can derive
$$
0<\omega<\frac{2 \pi}{3} \text {. }
$$

From (1), we have
$$
\sin \varphi=\frac{2}{A} \text {. }
$$

From (3), we have
$$
\sin (3 \omega+\varphi)=-\frac{2}{A} \text {. }
$$

Since $A>0$, we get
$$
\sin (3 \omega+\varphi)=-\sin \varphi \text {. }
$$

Thus,
$$
3 \omega+\varphi=\pi-\varphi+2 k_{2} \pi \left(k_{2} \in \mathbf{Z}\right) \text {, }
$$
which simplifies to
$$
3 \omega=2 \pi-2 \varphi+2 k_{2} \pi \left(k_{2} \in \mathbf{Z}\right) \text {, }
$$
or
$$
\omega=\frac{2 \pi-2 \varphi+2 k_{2} \pi}{3} \left(k_{2} \in \mathbf{Z}\right) \text {, }
$$
or
$$
\omega=\frac{2 \pi(1+k_{2})-2 \varphi}{3} \left(k_{2} \in \mathbf{Z}\right) \text {. }
$$

Since $0<\omega<\frac{2 \pi}{3}$, we have
$$
0<\frac{2 \pi(1+k_{2})-2 \varphi}{3}<\frac{2 \pi}{3} \text {, }
$$
which simplifies to
$$
0<2 \pi(1+k_{2})-2 \varphi<2 \pi \text {, }
$$
or
$$
0<\pi(1+k_{2})-\varphi<\pi \text {, }
$$
or
$$
0<\varphi<\pi(1+k_{2}) \text {, }
$$
or
$$
0<\varphi<\pi \text {, }
$$
and
$$
\varphi<\pi(1+k_{2}) \text {. }
$$

Since $0<\varphi<\pi$, we have
$$
0<\varphi<\pi \text {, }
$$
and
$$
0<\varphi<\pi(1+k_{2}) \text {. }
$$

Thus, $k_{2}=0$, and
$$
\omega=\frac{2 \pi-2 \varphi}{3} \text {. }
$$

From (2), we have
$$
\frac{5}{2} \omega+\varphi=k_{1} \pi \left(k_{1} \in \mathbf{Z}\right) \text {, }
$$
which simplifies to
$$
\frac{5}{2} \left(\frac{2 \pi-2 \varphi}{3}\right)+\varphi=k_{1} \pi \left(k_{1} \in \mathbf{Z}\right) \text {, }
$$
or
$$
\frac{5 \pi-5 \varphi}{3}+\varphi=k_{1} \pi \left(k_{1} \in \mathbf{Z}\right) \text {, }
$$
or
$$
\frac{5 \pi-2 \varphi}{3}=k_{1} \pi \left(k_{1} \in \mathbf{Z}\right) \text {, }
$$
or
$$
5 \pi-2 \varphi=3 k_{1} \pi \left(k_{1} \in \mathbf{Z}\right) \text {, }
$$
or
$$
2 \varphi=5 \pi-3 k_{1} \pi \left(k_{1} \in \mathbf{Z}\right) \text {, }
$$
or
$$
\varphi=\frac{5 \pi-3 k_{1} \pi}{2} \left(k_{1} \in \mathbf{Z}\right) \text {, }
$$
or
$$
\varphi=\frac{\pi(5-3 k_{1})}{2} \left(k_{1} \in \mathbf{Z}\right) \text {. }
$$

Since $0<\varphi<\pi$, we have
$$
0<\frac{\pi(5-3 k_{1})}{2}<\pi \text {, }
$$
which simplifies to
$$
0<5-3 k_{1}<2 \text {, }
$$
or
$$
3<3 k_{1}<5 \text {, }
$$
or
$$
1<k_{1}<\frac{5}{3} \text {, }
$$
or
$$
k_{1}=1 \text {. }
$$

Thus,
$$
\varphi=\frac{\pi(5-3 \cdot 1)}{2}=\frac{\pi(5-3)}{2}=\frac{2 \pi}{2}=\pi \text {, }
$$
which contradicts $0<\varphi<\pi$.

Therefore, the only function that meets the conditions is $y=4 \sin \left(\frac{\pi}{3} x+\frac{\pi}{6}\right)$.