The maximum value of g(t) is g(2)= 1.

F(s)>= 1 is established in Shanghang.

a/x+xlnx & gt； = 1

a & gt=x-x^2lnx

Let h (x) = x-x 2 lnx.

h`(x)= 1-2xlnx-x

Let h ` (x) = 0 and x = 1.

H(x) is declining.

The maximum value of h(x) is h( 1)= 1.

∴a>； = 1

(1)f'(x)= 1/x-a, which is the decreasing function in the interval (1, +∞), that is, when x >: 1, f'(x).

So 1/x-a

1/x & lt； a

Get a> 1.

g(x)'=e^x-a

According to the meaning of the question, (1, +∞) should have a minimum value, that is, when x >; 1, g'(x)>0, used to increase the function, so:

e^x-a>； 0

e^x>； a

Namely: e>a.

So the range of a is: (1, e).

(2) g (x)' = e x-a, and the interval (-1, +∞) is monotone increasing function, that is, when x >；; -1, g' (x) >: 0 is an increasing function, so:

e^x-a>； 0

e^x>； a

e^x>； e^(- 1)>； a

Then: a <1/e.

At this time f'(x)= 1/x-a,

When 0

When e

When x>1/a > e，f' (x)

So there is only one zero point.