7、The right focus of the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ is $F$, and the left directrix is $l$. The ellipse $C$ has $F$ and $l$ as its focus and directrix, respectively. A line with a slope of 1 is drawn through $F$ and intersects the ellipse $C$ at points $A$ and $B$. If the center $P$ of the ellipse $C$ is inside the circle with $A B$ as its diameter, then the range of the eccentricity $e$ of the ellipse $C$ is $\qquad$
7. $(0, \sqrt{2-\sqrt{2}})$

Analysis: It is easy to know that the focus $F(4,0)$, the left directrix $l: x=-1$, and the equation of the ellipse is: $\frac{(x-t)^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $A\left(x_{A}, x_{A}-4\right), B\left(x_{B}, x_{B}-4\right)$; Substituting $y=x-4$ into the ellipse equation yields:
$$
\left(a^{2}+b^{2}\right) x^{2}-\left(8 a^{2}+2 t b^{2}\right) x+16 a^{2}+b^{2} t^{2}-a^{2} b^{2}=0
$$

Since point $P$ is inside the circle with $A B$ as its diameter, we have $P A \cdot P B=2 x_{A} x_{B}-(4+t)\left(x_{A}+x_{B}\right)+t^{2}+16<0$. Using Vieta's formulas, we get: $(t-4)^{2}\left(a^{2}+b^{2}\right)<2 a^{2} b^{2}$. Since $c=t-4$, substituting gives $c^{2}\left(a^{2}+b^{2}\right)<2 a^{2} b^{2}$, solving which yields: $e<\sqrt{2-\sqrt{2}}$.