Solution: f (x) = | x+ 1 |+| x- 1 |, and the image is as shown in the figure.

(1) has x0∈R, which makes f(x0)≤m hold.

That is, m≥f(x)min, f (x) min = 2 and m≥2.

(2) Any m ∈ [1, 2] has m? +2m-f(t)≤0.

Then m ∈ [1, 2], f(t)≥(m? +2m) Maximum capacity.

m？ +2m=(m+ 1)？ - 1，m = 2，(m？ +2m) Max =8

F(t)≥8, that is | t+ 1 |+t- 1 | ≥ 8, and the solution is t≥4 or t≤-4.