解:
(1)设$y - 3=\frac{k}{x + 2}.$
因为$x = 2$时,$y = 7,$所以$7 - 3=\frac{k}{2 + 2},$
即$4=\frac{k}{4},$解得$k = 16,$所以$y$与$x$的函数表达式为$y - 3=\frac{16}{x + 2},$即$y=\frac{16}{x + 2}+3.$
(2)设$y_1 = k_1x,$$y_2=\frac{k_2}{x},$则$y = k_1x-\frac{k_2}{x}.$
当$x = 1$时,$y = k_1 - k_2 = -2,$①
当$x = -2$时,$y = -2k_1-\frac{k_2}{-2} = -2,$即$-2k_1+\frac{k_2}{2} = -2,$②
由①得$k_1 = k_2 - 2,$将其代入②得:
$-2(k_2 - 2)+\frac{k_2}{2} = -2,$
$-2k_2 + 4+\frac{k_2}{2} = -2,$
$-4k_2 + 8 + k_2 = -4,$
$-3k_2 = -12,$
解得$k_2 = 4.$
将$k_2 = 4$代入$k_1 = k_2 - 2$得$k_1 = 2.$
所以$y = 2x-\frac{4}{x}.$
当$x = -1$时,$y = 2\times(-1)-\frac{4}{-1}=-2 + 4 = 2.$