7. Given the hyperbola $\Gamma$:
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$

with left and right foci $F_{1}$ and $F_{2}$, respectively. Suppose a line $l$ passing through point $F_{2}$ intersects the right branch of the hyperbola $\Gamma$ at points $A$ and $B$, and $\left|A F_{1}\right|=\left|F_{1} F_{2}\right|,\left|B F_{2}\right|=2\left|A F_{2}\right|$. Then the eccentricity of the hyperbola $\Gamma$ is . $\qquad$
7. $\frac{5}{3}$.

As shown in Figure 6, let the midpoint of $A F_{2}$ be $M$, and connect $F_{1} M$.
By the problem, we know $c^{2}=a^{2}+b^{2}, c>a$.
Since $\left|A F_{1}\right|=\left|F_{1} F_{2}\right|=2 c$, and $M$ is the midpoint of $A F_{2}$, we have $F_{1} M \perp A F_{2}$.
$$
\begin{array}{l}
\text { By }\left|A F_{1}\right|-\left|A F_{2}\right|=2 a \\
\Rightarrow\left|A F_{2}\right|=2 c-2 a \\
\Rightarrow\left|M F_{2}\right|=\frac{1}{2}\left|A F_{2}\right|=c-a .
\end{array}
$$

In $\triangle M F_{1} F_{2}$, we have
$$
\begin{array}{l}
\cos \angle M F_{2} F_{1}=\frac{\left|M F_{2}\right|}{\left|F_{1} F_{2}\right|}=\frac{c-a}{2 c}, \\
\left|B F_{2}\right|=2\left|A F_{2}\right|=4 c-4 a .
\end{array}
$$

Then $\left|B F_{1}\right|=2 a+\left|B F_{2}\right|=4 c-2 a$.
In $\triangle B F_{1} F_{2}$, we have
$$
\begin{array}{l}
\cos \angle B F_{2} F_{1}=\frac{\left|F_{1} F_{2}\right|^{2}+\left|B F_{2}\right|^{2}-\left|B F_{1}\right|^{2}}{2\left|F_{1} F_{2}\right|\left|B F_{2}\right|} \\
\quad=\frac{4 c^{2}+16(c-a)^{2}-4(2 c-a)^{2}}{2 \times 2 c \times 4(c-a)} \\
=\frac{c-3 a}{4 c} . \\
\text { Since } \angle B F_{2} F_{1}+\angle M F_{2} F_{1}=\pi \\
\Rightarrow \cos \angle B F_{2} F_{1}+\cos \angle M F_{2} F_{1}=0 \\
\Rightarrow \frac{c-a}{2 c}+\frac{c-3 a}{4 c}=0 \\
\Rightarrow(5 a-3 c) c=0 \\
\Rightarrow 5 a=3 c \\
\Rightarrow e=\frac{c}{a}=\frac{5}{3} .
\end{array}
$$