(1) f' (x) = e x-a, then f'(0)= 1-a=- 1, and a=2.

Let f' (x) = e x-2 = 0, and the only stationary point x=ln2.

And f "(x) = e× x, f" (LN2) = 2 > 0.

Then the minimum value of f(x) f(ln2) =2( 1-ln2), which is also the minimum value.

(2) If G (x) = EX-X 2, then G' (x) = e x-2x = f (x),

The minimum value of f(x), that is, the minimum value of g'(x) is 2 (1-LN2) >: 0,

Then g(x) increases monotonically, and when x >; 0, g (x) = e x-x 2 > g (0) =1,that is, x 2 < e x.

(3)c & gt； 0, remember that h (x) = ce x-x 2. The images showing ce x and x 2 in the figure have at most two intersections:

x 1 & lt； X0。 Then when x & gtX0 h(x)>0, that is, x 2 < ce X.