8. 25 Detailed Explanation: Let $C_{1}: \frac{x^{2}}{m+9}+\frac{y^{2}}{m}=1, C_{2}: \frac{x^{2}}{9-m}-\frac{y^{2}}{m}=1(0<m<9)$. Suppose the intersection point $P\left(x_{0}, y_{0}\right)$, then from the fact that point $P$ lies on $C_{1}, C_{2}$, we have $\frac{x_{0}^{2}}{m+9}+\frac{x_{0}^{2}}{9-m}=2 \Rightarrow x_{0}^{2}=9-\frac{m^{2}}{9}$, thus $y_{0}^{2}=\frac{m^{2}}{9}$, where $x_{0}^{2}, y_{0}^{2} \neq 0$, , therefore $x_{0}^{2}+y_{0}^{2}=9$. From the sketch, it is known that the region enclosed contains 25 lattice points.
8. Given that the ellipse $C_{1}$ and the hyperbola $C_{2}$ share the foci $F_{1}(3,0), F_{2}(-3,0)$, and that the minor axis and the imaginary axis coincide. Then the number of lattice points inside the region enclosed by the trajectory of the intersection points of $C_{1}$ and $C_{2}$ is $\qquad$ .