The equation about $x$, $7 x^{2}-(k+13) x + k^{2}-k-2=0$, has two real roots $x_{1}, x_{2}$, and $00$.

Since $0<x_{1}<1<x_{2}<2$, we have
$$
1<x_{1}+x_{2}=\frac{k+13}{7}<3,
$$
$$
0<x_{1} x_{2}=\frac{k^{2}-k-2}{7}<2 \text {. }
$$
Solve (1), (2), and (3) to get:
$$
\frac{3-2 \sqrt{21}}{3}0,
\end{array}
$$

The graph opens upwards, because
$$
f\left(x_{1}\right)=f\left(x_{2}\right)=0
$$
(Figure 2), hence \( f(0) > 0 \) and \( f(1) < 0 \), which leads to the system of inequalities:
$$
\left\{\begin{array}{l}
k^{2}-k-2>0, \\
k^{2}-2k-80 < 0.
\end{array}\right.
$$

Solving this yields \( -2 < k < -1 \) or \( 3 < k < 4 \), which is correct and has a general nature. It is not difficult to prove the following theorem:
If \( f(x) \) is a quadratic function (continuous function), and \( a < b \), if \( f(a) \) and \( f(b) \) have opposite signs, then there must be \( x_{0} \in (a, b) \) such that \( f\left(x_{0}\right)=0 \).
(Author's affiliation: Tianjin Baodi County High School)