(I)f′(x)= 2x+xcos x，

The curve y=f(x) is tangent to the straight line y=b at points (a, f(a)).

∴f'(a)= 0, f(a)=b, simultaneous.

2a+acosa=0

a^2+asina+cosa=b

The solution is a=0 b= 1.

Therefore, a=0 and b = 1.

(II)∫f′(x)= x(2+cosx)。

So when x > 0, f' (x) > 0, so f(x) increases monotonously.

When x < 0, f ′ (x) < 0, and f(x) decreases monotonically.

When x=0, f(x) takes the minimum value f(0)= 1,

So when b > 1, the curve y=f(x) and the straight line y=b have two different intersections, so the value range of b is (1, +∞).

Hope to adopt, thank you.