f'(x)=a( 1-2/x)- 1/x^2+2/x^3

=( 1-2/x)(a- 1/x^2)

=(x-2)(ax^2- 1)/x^3.

( 1)a & lt； =0 ax 2- 1

X>2 When f' (x) < 0, f(x) is a decreasing function.

A> is at 0, ax 2-1= a (x+1√ a) (x-1√ a), which is known by ordinal axis labeling method.

I) x > when a = 1/4; 2 hours f'(x)>0, f(x) is increasing function; 0<x<2 when f' (x) < 0, f(x) is a decreasing function;

ii)0 & lt； a & lt0 < at 1/4； X<2 or x & gtF' (x) is in1√ a >: 0, and f(x) is increasing function.

2<x & ltF' (x) is in1√ a < 0, and f(x) is a decreasing function;

iii)a & gt； 0 < at 1/4； X<1√ a or x>2 hours f'(x)>0, f(x) is increasing function,

1/√a & lt； X<2 when f' (x) < 0, f(x) is a decreasing function.

(2)f(x) has two zeros:

I)a & lt； When =0, f (2) = a (2-2LN2)+1/4 >; 0,

so- 1/(8-8ln 2)< a & lt； =0;

ii)a & gt； When f(2)> 0; 0,

f( 1/√a)= a( 1/√a+lna)+√a-a & lt； 0,

That is 2+√ ALNA-√ A.

Let u=√a, g(u)=2+2ulnu-u,

g'(u)=2lnu+ 1=0，u0= 1/√e，

g(u)>= g(u0)= 2-2/√e & gt； 0,

So ① there is no solution, and f(x) has no two zeros.

To sum up,-1/(8-8ln2) < a <; =0。