7. Let the incircle radius of equilateral $\triangle A B C$ be 2, and the incenter be
I. If point $P$ satisfies $P I=1$, then the maximum value of the ratio of the areas of $\triangle A P B$ to $\triangle A P C$ is $\qquad$.
7. $\frac{3+\sqrt{5}}{2}$.

As shown in Figure 4, given $P I=1$, we know that point $P$ lies on the unit circle $\odot I$.

Let $\angle B A P=\alpha$. On $\odot I$, take a point $P_{0}$ such that $\alpha$ reaches its maximum value $\alpha_{0}$. At this time, point $P_{0}$ should fall within $\angle I A C$, and it is the point where $A P_{0}$ is tangent to $\odot I$.
Since $0<\alpha \leqslant \alpha_{0}<\frac{\pi}{3}$, we have
$$
\begin{array}{l}
\frac{S_{\triangle A P B}}{S_{\triangle A P C}}=\frac{\frac{1}{2} A P \cdot A B \sin \alpha}{\frac{1}{2} A P \cdot A C \sin \left(\frac{\pi}{3}-\alpha\right)}=\frac{\sin \alpha}{\sin \left(\frac{\pi}{3}-\alpha\right)} \\
\leqslant \frac{\sin \alpha_{0}}{\sin \left(\frac{\pi}{3}-\alpha_{0}\right)}=\frac{\sin \left(\frac{\pi}{6}+\theta\right)}{\sin \left(\frac{\pi}{6}-\theta\right)},
\end{array}
$$

where $\theta=\alpha_{0}-\frac{\pi}{6}=\angle I A P_{0}$.
Since $\angle A P_{0} I=\frac{\pi}{2}$, we know $\sin \theta=\frac{I P_{0}}{A I}=\frac{1}{2 r}=\frac{1}{4}$.
Thus, $\cot \theta=\sqrt{15}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{\sin \left(\frac{\pi}{6}+\theta\right)}{\sin \left(\frac{\pi}{6}-\theta\right)}=\frac{\frac{1}{2} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta}{\frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta} \\
=\frac{\cot \theta+\sqrt{3}}{\cot \theta-\sqrt{3}}=\frac{\sqrt{15}+\sqrt{3}}{\sqrt{15}-\sqrt{3}}=\frac{3+\sqrt{5}}{2} .
\end{array}
$$

According to equations (1) and (2), when point $P$ coincides with $P_{0}$, the maximum value of $\frac{S_{\triangle A P B}}{S_{\triangle A P C}}$ is $\frac{3+\sqrt{5}}{2}$.