China Mathematical Competition

By symmetry, suppose the equation of $\Gamma$ is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$), and $A(a, 0)$, $B(0, b)$, $F_1(-c, 0)$, $F_2(c, 0)$, where $c = \sqrt{a^2 - b^2}$.
By the given conditions, we know that
$$\begin{aligned} \overrightarrow{AF_1} \cdot \overrightarrow{AF_2} + \overrightarrow{BF_1} \cdot \overrightarrow{BF_2} &= (-c-a)(c-a) + (-c^2+b^2) \\ &= a^2 + b^2 - 2c^2 = 0. \end{aligned}$$
Thus, $a^2 + b^2 - 2c^2 = -a^2 + 3b^2 = 0$, and hence $a = \sqrt{3}b, c = \sqrt{2}b$.
Let $O$ be the origin of the coordinates, and then
$$\begin{aligned} \tan \angle ABO &= \frac{a}{b} = \sqrt{3}, \\ \tan \angle OBF_1 &= \tan \angle OBF_2 = \frac{c}{b} = \sqrt{2}. \end{aligned}$$
Therefore,
$$$$
$$\begin{aligned} & \tan \angle ABF_1 \cdot \tan \angle ABF_2 \\ &= \tan(\angle ABO + \angle OBF_1) \cdot \tan(\angle ABO - \angle OBF_1) \\ &= \frac{\sqrt{3} + \sqrt{2}}{1 - \sqrt{3} \cdot \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{1 + \sqrt{3} \cdot \sqrt{2}} = -\frac{1}{5}. \end{aligned}$$