8. Given complex numbers $z_{1}, z_{2}$ satisfy: $\left|z_{1}\right|=1,\left|z_{2}\right|=1$, and $z_{1}+z_{2}=-\frac{7}{5}+\frac{1}{5} \mathrm{i}$, then $z_{1} z_{2}=$ $\qquad$
8. $\frac{24}{25}-\frac{7}{25} \mathrm{i}$ Let $z_{1}=\cos \theta_{1}+\mathrm{i} \sin \theta_{1}, z_{2}=\cos \theta_{2}+i \sin \theta_{2}$

So $\left\{\begin{array}{l}\cos \theta_{1}+\cos \theta_{2}=-\frac{7}{5} \\ \sin \theta_{1}+\sin \theta_{2}=\frac{1}{5}\end{array}\right.$ Therefore, $z_{1} \times z_{2}=\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)$
$$
\left\{\begin{array}{l}
2 \cos \frac{\theta_{1}+\theta_{2}}{2} \cos \frac{\theta_{1}-\theta_{2}}{2}=-\frac{7}{5} \\
2 \sin \frac{\theta_{1}+\theta_{2}}{2} \cos \frac{\theta_{1}-\theta_{2}}{2}=\frac{1}{5}
\end{array}\right.
$$

So $\tan \frac{\theta_{1}+\theta_{2}}{2}=-\frac{1}{7} \quad$ Therefore, $\cos \left(\theta_{1}+\theta_{2}\right)=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{24}{25}, \sin \left(\theta_{1}+\theta_{2}\right)=\frac{-\frac{2}{7}}{1+\frac{1}{49}}=-\frac{7}{25}$ Therefore, $z_{1} \times z_{2}=\frac{24}{25}-\frac{7}{25} \mathrm{i}$.