7. Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with an eccentricity of $\frac{\sqrt{5}}{2}, F_{1}, F_{2}$ are the left and right foci of $C$, respectively. A line $l$ passing through $F_{2}$ intersects $C$ at points $A$ and $B$ (point $A$ is in the first quadrant), and $\overrightarrow{A F_{2}}=3 \overrightarrow{F_{2} B}$. If the area of $\triangle F_{1} A B$ is $\frac{32}{3}$, then the inradius of $\triangle F_{1} A B$ is $\qquad$.
Solving
By the focal radius formula $\left|F_{2} A\right|=\frac{e p}{1-e \cos \alpha},\left|F_{2} B\right|=\frac{e p}{1+e \cos \alpha} \Rightarrow e \cos \alpha=\frac{1}{2}$,
thus $\cos \alpha=\frac{\sqrt{5}}{5} \Rightarrow \tan \alpha=2$. Let $A B: y=2(x-\sqrt{5} b), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
solving $\left\{\begin{array}{l}y=2(x-\sqrt{5} b), \\ x^{2}-4 y^{2}=4 b^{2}\end{array} \Rightarrow 15 x^{2}-32 \sqrt{5} b x+84 b^{2}=0\right.$.
Then $\left|x_{1}-x_{2}\right|=\frac{\sqrt{32^{2} \cdot 5 b^{2}-4 \cdot 15 \cdot 84 b^{2}}}{15}=\frac{4 \sqrt{5} b}{15}$, hence $|A B|=\sqrt{5}\left|x_{1}-x_{2}\right|=\frac{4}{3} b$.
Also, $S_{\triangle F_{1} A B}=\sqrt{5} b \cdot\left|y_{1}-y_{2}\right|=\sqrt{5} b \cdot \frac{8 \sqrt{5} b}{15}=\frac{8 b^{2}}{3}=\frac{32}{3} \Rightarrow b=2, a=4$,
so $2 s_{\triangle F_{1} A B}=4 a+2|A B|=16+\frac{16}{3}=\frac{64}{3} \Rightarrow r=\frac{S_{\triangle F_{1} A B}}{s_{\triangle F_{1} A B}}=1$.