2. Let the real-coefficient quadratic equation $x^{2}+a x+2 b-$ $2=0$ have two distinct real roots, one of which lies in the interval $(0,1)$, and the other in the interval $(1,2)$. Then the range of $\frac{b-4}{a-1}$ is . $\qquad$
2. $\left(\frac{1}{2}, \frac{3}{2}\right)$.

Solution: According to the problem, let the two distinct real roots be $x_{1}, x_{2}$, and $0 < x_{1} < 1 < x_{2} < 2$, then $1 < x_{1} + x_{2} = -a < 3, 0 < x_{1} x_{2} = 2b - 2 < 2$.

Thus, $-3 < a < -1, 1 < b < 2$, which means $-\frac{1}{2} < \frac{1}{a-1} < -\frac{1}{4}, -3 < b-4 < -2$.
Therefore, $\frac{1}{2} < \frac{b-4}{a-1} < \frac{3}{2}$, so the range of values is $\left(\frac{1}{2}, \frac{3}{2}\right)$.