7. Given the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ with left and right foci $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{1}$ intersects the left branch at points $A$ and $B$, and $\frac{\left|A F_{1}\right|}{\left|A F_{2}\right|}=\frac{1}{2}$. Then $\triangle A F_{1} F_{2}$ and $\triangle A B F_{2}$
7. 4:9.

As shown in Figure 3, let the incircles of $\triangle A F_{1} F_{2}$ and $\triangle A B F_{2}$ be $\odot J$ and $\odot I$, respectively, and $A I$ intersects the $x$-axis at point $D$.
According to the problem, we know that $\left|A F_{1}\right|=6$ and $\left|A F_{2}\right|=12$.
From $\left|A F_{2}\right|=12=3-\frac{5}{3} x_{A}$
$\Rightarrow A\left(-\frac{27}{5}, \frac{8 \sqrt{14}}{5}\right)$.
Also, point $D\left(-\frac{5}{3}, 0\right)$, thus, the equation of the line $A D$, which is parallel to $\angle F_{1} A F_{2}$, is
$$
x=-\frac{\sqrt{14}}{6} y-\frac{5}{3} \text {. }
$$

By the property, we know that point $J\left(-3, \frac{8 \sqrt{14}}{14}\right)$.
Therefore, the inradius of $\triangle A F_{1} F_{2}$ is $r_{1}=\frac{8 \sqrt{14}}{14}$.
Let $\angle F_{1} A F_{2}=\theta$.
Hence, $\cos \theta=\frac{6^{2}+12^{2}-10^{2}}{2 \times 6 \times 12}=\frac{5}{9}$
$\Rightarrow \tan \frac{\theta}{2}=\sqrt{\frac{2}{7}}$.
Thus, the inradius of $\triangle A B F_{2}$ is
$r_{2}=\left|A F_{1}\right| \tan \frac{\theta}{2}=\frac{6 \sqrt{14}}{7}$.
Therefore, the required ratio is $4: 9$.