8. The range of the function $y=\frac{x+1}{x-1}$ on the interval $(-\infty, 0) \cup[2,5)$ is $\qquad$
8. $\left(\frac{3}{2}, 3\right] \cup(-1,1) \cdot y=1+\frac{2}{x-1}, x-1 \in(-\infty,-1) \cup[1,4), \frac{1}{4}<\frac{1}{x-1} \leqslant 1$ or $-1<\frac{1}{x-1}<$
0. Thus $y \in\left(\frac{3}{2}, 3\right] \cup(-1,1)$.