7. Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have its left focus at point $F$, and let a tangent line be drawn from a point $A$ on the ellipse $C$ intersecting the $y$-axis at point $Q$. If $\angle Q F O=45^{\circ}, \angle Q F A=30^{\circ}$, then the eccentricity of the ellipse is $\qquad$.
7. $\frac{\sqrt{6}}{3}$.

As shown in Figure 2, let $A\left(x_{0}, y_{0}\right)$, then the equation of the tangent line passing through point $A$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$. Let the tangent line intersect the $x$-axis at point $P$, then $P\left(\frac{a^{2}}{x_{0}}, 0\right), C\left(0, \frac{b^{2}}{y_{0}}\right)$, so
$$
\begin{array}{l}
\overrightarrow{F Q} \cdot \overrightarrow{F O}=\left(c, \frac{b^{2}}{y_{0}}\right) \cdot(c, 0)=c^{2}, \\
\overrightarrow{F Q} \cdot \overrightarrow{F A}=\left(c, \frac{b^{2}}{y_{0}}\right) \cdot\left(x_{0}+c, y_{0}\right)=c x_{0}+a^{2} .
\end{array}
$$

Thus,
$$
\begin{aligned}
\frac{\cos \angle Q F O}{\cos \angle Q F A} & =\frac{\overrightarrow{F Q} \cdot \overrightarrow{F A}}{\overrightarrow{F Q} \cdot \overrightarrow{F O}} \cdot \frac{|\overrightarrow{F Q}| \cdot|\overrightarrow{F O}|}{|\overrightarrow{F Q}| \cdot|\overrightarrow{F A}|} \\
& =\frac{c^{2}}{c x_{0}+a^{2}} \cdot \frac{e x_{0}+a}{c} \\
& =\frac{c}{a} \cdot \frac{e x_{0}+a}{e x_{0}+a}=e,
\end{aligned}
$$

Therefore, the eccentricity of the ellipse
$$
e=\frac{\cos \angle Q F O}{\cos \angle Q F A}=\frac{\cos 45^{\circ}}{\cos 30^{\circ}}=\frac{\sqrt{6}}{3} .
$$