已知函数f(x)对任意实数的a,b∈R满足:f(a+b)=f(a)+f(b)-6,当a>0时,f(a)

(1)令a=-2,b=0.
有f(-2)=f(-2)+f(0)-6,
所以f(0)=6
令a=2,b=-2.
f(0)=f(2)+f(-2)-6,
所以f(2)=0.
(2)设X1>X2,
f(X1)=f(X1-X2)+f(X2)-6
f(X1)-f(X2)=f(X1-X2)-6