解:
(1) 将原方程组的两个方程相加,得$3(x + y) = 6m + 1,$
把$x + y = 1$代入,得$6m + 1 = 3,$解得$m = \frac{1}{3}。$
(2) 将原方程组的两个方程相减,得$x - y = 2m - 1,$
因为$-1 ≤ x - y ≤ 5,$所以$-1 ≤ 2m - 1 ≤ 5,$
不等式两边同时加1,得$0 ≤ 2m ≤ 6,$
两边同时除以2,得$0 ≤ m ≤ 3。$
(3) 当$0 ≤ m ≤ \frac{3}{2}$时,
$|m + 2| + |2m - 3| = (m + 2) - (2m - 3) = 5 - m;$
当$\frac{3}{2} < m ≤ 3$时,
$|m + 2| + |2m - 3| = (m + 2) + (2m - 3) = 3m - 1。$