f'(x)=(3x^2+6x-3)e^(-x)-(x^3+3x^2-3x-3)e^(-x)

=(-x^3+9x)e^(-x)=0

Solution: x 1=-3, x2=0, x3=3.

F"(x)=(

x^3-3x^2-9x+9)e^(-x)

F"(x 1)=- 18

e^3<； 0,

f "(x2)= 9 & gt； 0,

F"(x3)=- 18e^(-3)

The function f(x) obtains the maximum, minimum and maximum at x 1, x2 and x3, respectively.

∴ function f(x)

In (-∞, -3), (0, 3) monotonically increasing; Monotonically decreasing at (-3,0), (3, +∞);

(2)∵ function f (x) = (x 3+3x 2+ax+b) e (-x)

f'(x)=(3x^2+6x+a-x^3-3x^2-ax-b)e^(-x)

=(-x^3+(6-a)x+a-b)e^(-x)

The function f(x) monotonically increases at (-∞, α), (2, β) and monotonically decreases at (α, 2), (β, +∞).

∴x=2 is the minimum point of the function f(x).

That is f' (2) = (-8+12-2a+a-b) e (-2) = 0 = >; a+b=4

∴b=4-a

Substitute-x3+(6-a) x+a-b =-x3+(6-a) x+2a-4 = (x-2) (-x2-2x+2-a).

∴ α, β satisfies-x 2-2x+2-a = 0 = > x 2+2x+a-2 = 0.

Vieta's theorem knows that β+α=-2,

βα=a-2

(β-α)^2=(β+α)^2-

4βα= 12-4a

β-α=2√(3-a)

When a <-6, when a+b=4, it satisfies β-α >; six

So the second problem of this problem is problematic, that is, β-α > has never been satisfied; six