7. Let the incircle of equilateral triangle $ABC$ have a radius of 2, with the center at $I$. If point $P$ satisfies $PI=1$, then the maximum value of the ratio of the areas of $\triangle APB$ to $\triangle APC$ is $\qquad$

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Given $\angle P A C=\theta$, then $\frac{S_{\triangle A P B}}{S_{\triangle A P C}}=\frac{\sin \left(60^{\circ}-\theta\right)}{\sin \theta}$ $=\frac{\sqrt{3}}{2} \cot \theta-\frac{1}{2}$. Therefore, when $\theta$ is the smallest, $\frac{S_{\triangle A P B}}{S_{\triangle A P C}}$ reaches its maximum value, at which point $A P$ is tangent to the circle $I$ with a radius of 1.
Let $\angle P A I=\varphi, A I=4 \Rightarrow \tan \varphi=\frac{1}{\sqrt{15}}$
$$
\begin{array}{l}
\Rightarrow \tan \theta=\tan \left(30^{\circ}-\varphi\right)=\frac{\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{15}}}{1+\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{15}}}=\frac{\sqrt{15}-\sqrt{3}}{3 \sqrt{5}+1}=\frac{\sqrt{3}(\sqrt{5}-1)}{3 \sqrt{5}+1} \\
\Rightarrow \cot \theta=\frac{3 \sqrt{5}+1}{\sqrt{3}(\sqrt{5}-1)}=\frac{4+\sqrt{5}}{\sqrt{3}} \Rightarrow \frac{\sqrt{3}}{2} \cot \theta-\frac{1}{2}=\frac{3+\sqrt{5}}{2} .
\end{array}
$$

In summary, the maximum value of the ratio of the areas of $\triangle A P B$ to $\triangle A P C$ is $\frac{3+\sqrt{5}}{2}$.