(1) let X 1 >: x2.

f(x)+f(y)=f(x+y)

Let x=x2 and x+y=x 1,

y = x 1-x2 >； 0

So f (x2)+f (x1-x2) = f (x1).

So f (x1)-f (x2) = f (x1-x2) < 0.

Therefore, f(x) is a decreasing function on R.

(2)f(x)+f(y)=f(x+y)

Let x=y=0, then f (0)+f (0) = f (0), and f (0) = 0.

f(3)= f(2)+f( 1)= f( 1)+f( 1)+f( 1)=-2

f(3)+f(-3)=f(0)=0

f(-3)=2

F(x) is a decreasing function at-3,3,

Therefore, the maximum value is f(-3)=2.

The minimum value is f(3)=-2.