When (1) a = 0, f( 1) =-1.

f '(x)=[(x- 1)^2+2(x- 1)]e^x-[( 1/x)x-( 1+lnx)]/x^2

=(x- 1)(x+ 1)e^x +lnx/x^2

F'( 1) = 0, and the tangent equation at point (1, f( 1)) is y =-1.

(2)a & gt； 0,

The stagnation point of f' (x) = (x-1) (x+1) ex+lnx/x2x =1

f ' '(x)=(x^2+2x- 1)e^x+( 1-2lnx)/x^3，f ' '( 1)= 2e+ 1 & gt； 0,

X = 1 is the minimum point, that is, the minimum point.

When 0

lim & ltx→0+>f(x)= lim & lt； x→0+>[(x- 1)^2 e^x-( 1+lnx)/x+a]=+∞

lim & ltx→+∞& gt； f(x)= lim & lt； x→+∞& gt； [(x- 1)^2 e^x-( 1+lnx)/x+a]=+∞

At this time, f(x) has two zeros;

When a = 1, f( 1) = a- 1 = 0, then f(x) has 1 zeros, and x =1;

When a > is in 1, f (1) = a-1>; 0, when f(x) is not zero.