8. $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$.

Let the equation of the hyperbola $C$ be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with the semi-focal distance $c=\sqrt{a^{2}+b^{2}}$. Denote the points where the incircle of $\triangle A F_{1} F_{2}$ touches the $x$-axis, $A F_{1}$, and $A F_{2}$ as $E, P, Q$, respectively. Then, $A P=A Q, F_{1} E=F_{1} P, F_{2} E=F_{2} Q$, so $A F_{1}-A F_{2}=2 a, F_{1} F_{2}=2 c$. Therefore, $F_{1} E=c+a, F_{2} E=c-a$, which means point $E$ is the right vertex of the hyperbola.

Denote the points where the incircle of $\triangle B F_{1} F_{2}$ touches the $x$-axis, $B F_{1}$, and $B F_{2}$ as $E^{\prime}, R, S$, respectively. Similarly, $E^{\prime}$ is the right vertex of the hyperbola, i.e., $E=E^{\prime}$, thus $I_{1} I_{2} \perp x$-axis.

Given the eccentricity of the hyperbola $C$ is $\frac{c}{a}=2$, we have $c=2 a$. Therefore, $F_{2} Q=F_{2} S=F_{2} E=c-a=a$. Connecting $I_{1} Q, I_{2} S$, in the right trapezoid $I_{1} I_{2} S Q$, $\angle Q I_{1} I_{2}$ equals the inclination angle of line $l$, i.e., $\sin \angle Q I_{1} I_{2}=\frac{8}{9}$, $2 a=Q S=I_{1} I_{2} \sin \angle Q I_{1} I_{2}=\frac{9}{2} \times \frac{8}{9}=4$, so $a=2, c=4, b=\sqrt{c^{2}-a^{2}}=2 \sqrt{3}$. Therefore, the equation of the hyperbola $C$ is $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$.
8. Given points $F_{1}, F_{2}$ are the left and right foci of a hyperbola $C$ centered at the origin with the coordinate axes as its axes of symmetry. A line $l$ passing through point $F_{2}$ intersects the right branch of the hyperbola at points $A, B$. Points $I_{1}, I_{2}$ are the incenter of $\triangle A F_{1} F_{2}$ and $\triangle B F_{1} F_{2}$, respectively. If the eccentricity of the hyperbola $C$ is $2$, $\left|I_{1} I_{2}\right|=\frac{9}{2}$, and the sine of the inclination angle of line $l$ is $\frac{8}{9}$, then the equation of the hyperbola $C$ is . $\qquad$