(1)∵|x-2|+|x-1|≥|(x-2)-(x-1)|=1恒成立,当x≤1或x≥2时,f(x)取最小值1,
∴由题意得,m=1.
(2)∵ |x-3|-|x-1|≤1,
∴可分类讨论或画数轴解得x≥3/2.