When f' (x) < 0, x < 0；; When f' (x) > 0, x > 0.

So the subtraction interval of the function f(x) is (-∞, 0); The increased interval is (0, +∞)

(2) Proof: (i) g (x) = f' (x) = ex (x-a+1)+(a-1), g'(x)=ex(x-a+2.

When g' (x) < 0, x < a-2; When g' (x) > 0, x > a-2.

Because a > 2, the function g(x) decreases on (0, a-2); Increment on (a-2, +∞)

And because g(0)=0 and g (a) = ea+a- 1 > 0,

So there is only one x0 on (0, +∞), which makes g (x0) = 0.

(2) Solution: If a≤2, f(x) can be monotonically increased in [0,2] when x ∈ [0 0,2] and g(x)≥0, and f(0)=0.

∴f(x)≥f(0)=0, which is irrelevant.

∴a>2

According to (i), f(x) decreases at (0, x0) and increases at (x0, +∞).

Let the maximum value of f(x) on [0,2] be m, then M=max{f(0), f(2)},

If f(x)≤0 holds for any x ∈ [0 0,2], then f(0)≤0 f(2)≤0.

F(2)≤0+2a-2+a ≤ 0 (2-a) E2, ∴ a ≥ 2e2-2e2-3 = 2+4e2-3 > 2,

F(0)=0，∴a≥2e2-2 /e2-3。 Can you understand? Welcome to continue your inquiry.