Example 4 The equation $7 x^{2}-(k+13) x+k^{2}-k-2$ $=0$ (where $k$ is a constant) has two real roots $\alpha, \beta$, and $0<\alpha<1$, $1<\beta<2$. Then, the range of values for $k$ is ( ).
(A) $3<k<4$
(B) $-2<k<-1$
(C) $-2<k<-1$ or $3<k<4$
(D) No solution
(1990, National Junior High School Mathematics League)
Solution: Let $f(x)=7 x^{2}-(k+13) x+k^{2}-k-2$.
Since $7>0$, the graph of $f(x)$ opens upwards. To make the two real roots $\alpha$ and $\beta$ satisfy $0<\alpha<1$ and $0<\beta<1$, we need:
$$
\left\{\begin{array}{l}
\Delta=(k+13)^{2}-4 \cdot 7 \cdot\left(k^{2}-k-2\right)>0, \\
0<\frac{k+13}{14}<1, \\
f(0)=k^{2}-k-2>0, \\
f(1)=7-(k+13)+k^{2}-k-20
\end{array}\right.
$$

Solving (1) gives $k>2$ or $k3$ or $k<0$.
Therefore, the solution to the system of inequalities is $-2<k<-1$ or $3<k<4$. Hence, the correct choice is (C).

Note: According to the characteristics of the parabola's graph, combined with the discriminant, axis of symmetry, and endpoint values to determine the distribution of the roots of a quadratic equation, this is a commonly used method.