|Column 15 Given the sets $A=\left\{x \mid x^{2}-x-2<0\right\}, B=\left\{x \mid x^{2}-a x-a^{2}<0\right.$, $a \in \mathbf{R}\}$. If $A \subseteq B$, find the range of real numbers for $a$.
Solve $A=(-1,2)$, let $f(x)=x^{2}-a x-a^{2}$. Since $A \subseteq B$, when $x \in(-1, 2)$, $f(x) < 0$. 

We have:
$$
\left\{\begin{array}{l}
f(-1) \leqslant 0, \\
f(2) \leqslant 0, \\
-\frac{a}{2} \leqslant-1 \text{ or }-\frac{a}{2} \geqslant 2, \\
\left\{\begin{array}{l}
-\frac{a}{2} \geqslant-1, \\
-\frac{a}{2} \leqslant 2, \\
\Delta=a^{2}+4 a^{2} > 0 .
\end{array}\right.\right.\right.
$$

Solving the inequalities, we get:
$$
\left\{\begin{array}{l}
1+a-a^{2} \leqslant 0, \\
4-2 a-a^{2} \leqslant 0, \\
a \leqslant-2 \text{ or } a \geqslant 4, \\
-4 \leqslant a \leqslant 2, \\
a \neq 0 .
\end{array}\right.
$$

From the above, we have:
$$
\left\{\begin{array}{l}
a \leqslant \frac{1-\sqrt{5}}{2} \text{ or } a \geqslant \frac{1+\sqrt{5}}{2}, \\
a \leqslant-1-\sqrt{5} \text{ or } a \geqslant \frac{1+\sqrt{5}}{2}, \\
a \leqslant-2 \text{ or } a \geqslant 4, \\
-4 \leqslant a \leqslant 2, \\
a \neq 0 .
\end{array}\right.
$$

Therefore, the range of $a$ is $(-\infty,-1-\sqrt{5}) \cup\left(\frac{1+\sqrt{5}}{2},+\infty\right)$.