(1), where f(xy)=f(x)+f(y)

Let x=y= 1, then f (1) = f (1)+f (1), that is, f( 1)=2f( 1).

∴ f( 1)=0

(2), f(x) is a function defined on (0, +∞)

∴ x>0，2-x>0

∴ x∈(0，2)

According to the meaning of the question, f(x)+f(2-x)=fx(2-x)

If f( 1/3)= 1, then f (1/9) = f (1/3)+f (1/3) = 2.

The original inequality is simplified as FX (2-x) < f (1/9).

∫f(x) is a decreasing function on x∈(0, +∞).

∴ x(2-x)> 1/9

Simplified to 9x? - 18x+ 1 x 1。

f(x2)-f(x 1)= f(x2)+f(-x 1)= f(x2-x 1)

From x2 > x 1, x2-x 1 > 0 is obtained.

If x > 0, f (x) < 0.

∴f(x2)-f(x 1)= f(x2-x 1)< 0

The function f(x) is a monotonically decreasing function.

(3), ∫f(x) is a monotonically decreasing function on the real number r.

∴ when x∈- 12, 12, f(x)max=f(- 12), f(x)min=f( 12).

First calculate f( 12)

f( 12)= f(6+6)= f(6)+f(6)= 2f(6)= 2f(3)+f(3)= 4f(3)=-8

f(- 12)=-f( 12)=8

∴ when x∑- 12, 12, f(x)max=8, f(x)min=-8.

3 ,

f(x)=x？ +ax+3=(x+a/2)？ +3-a？ /4 (x∈-2，2)

According to the position of the symmetry axis x=-a/2, three cases are discussed:

1-A/2 4。

f(x)min=f(-2)=7-2a

In order to make f(x)≥a constant, 7-2a≥a and a≤7/3.

Consider a > 4, and know that A does not exist.

2-2 ≤-A/2 ≤ 2, that is, a ∈-4,4.

f(x)min=f(-a/2)=3-a？ /4

In order to make f(x)≥ a constant, then 3-a? /4≥a

Simplify, get an A? +4a- 12≤0

Solve the inequality and get a ∈-6,2.

Consider a ∑-4,4

Get a ∑-4,2.

3-A/2 > 2, that is, A.

f(x)min=f(2)=7+2a

In order to make f(x)≥a constant, 7+2a≥a, and a≥-7 is obtained.

Consider a < -4 and get a∈-7, -4)

Take the union of the above three situations, and the value range of a is a ∈-7,2.