8. If the complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3,3 z_{1}-2 z_{2}=2-l$, then $z_{1} z_{2}=$
$$
-\frac{30}{13}+\frac{72}{13} i
$$
8. [Analysis and Solution] The reference solution is quite skillful, but based on the characteristics of the problem, using the trigonometric form of complex numbers seems more in line with students' thinking patterns and is not complicated.

Let $z_{1}=2(\cos \alpha+i \sin \alpha), z_{2}=3(\cos \beta+i \sin \beta)$, then from $3 z_{1}-2 z_{2}=\frac{3}{2}-i$ and the necessary and sufficient conditions for the equality of complex numbers, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
6(\cos \alpha-\cos \beta)=\frac{3}{2} \\
6(\sin \alpha-\sin \beta)=-1
\end{array}\right. \\
\left\{\begin{array}{l}
-12 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)=\frac{3}{2} \\
12 \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)=-1
\end{array},\right.
\end{array}
$$

Dividing the two equations, we get $\tan \left(\frac{\alpha+\beta}{2}\right)=\frac{3}{2}$,
By the universal formula, we get
$$
\begin{aligned}
\sin (\alpha+\beta) & =\frac{12}{13}, \cos (\alpha+\beta)=-\frac{5}{13} . \\
\text { Hence } z_{1} \cdot z_{2} & =6[\cos (\alpha+\beta)+i \sin (\alpha+\beta)] \\
& =-\frac{30}{13}+\frac{72}{13} i .
\end{aligned}
$$