f(x)=x^3+ax^2+x+ 1,f'(x)=3x^2+2ax+ 1

When 4a 2- 12 ≤ 0, that is, -√3≤a≤√3, F' (x) > 0 holds.

F(x) monotonically increases in (-∞,+∞).

When 4a 2- 12 >: 0, that is, when a≤-√3 or a≥√3, f'(x)=0 has two real number solutions.

Note that x 1 = [-a-√ (A 2-3)]/3, x2 = [-a+√ (A 2-3)]/3,

F(x) monotonically increases in (-∞, x 1), monotonically decreases in [x 1, x2] and monotonically increases in [x2,+∞).

2.

Because f(x) is a decreasing function in the interval (-2/3,-1/3).

So f' (x) = 3x 2+2ax+ 1 ≤ 0.

So f' (-2/3) < 0, f'(- 1/3)≤0.

So a≥2

Another method

F(x) decreases in the interval (-2/3,-1/3), where f' (x)

This interval is contained in [x 1, x2].

X 1 = [-a-√ (A 2-3)]/3 ≤-2/3，x2 = [-a+√ (A 2-3)]/3 ≥- 1/3，

The solution is a≥7/4 and a≥2, that is, a≥2.

The range of a [2, +∞).