(1)令x=0,y=1,则f(0+1)=f(0)f(1),
∵当x>0时,f(x)>1,∴f(1)>1,∴f(0)=1;
(2)证明:设x 1<x 2,则x 2-x 1>0
∵当x>0时,f(x)>1,∴f(x 2-x 1)>1
∴f(x 2)=f(x 2-x 1+x 1)=f(x 2-x 1)f(x 1)>f(x 1)
∴f(x)在R上是增函数;
(3)∵f(x)在R上是增函数,f(k•3 x) f(3 x-9 x-2)=f(k 3 x+3 x-9 x-2)<f(0),
∴3 2x-(1+k)•3 x+2>0对任意x∈R成立.
∴1+k<3 x+
2
3 x
∵3 x>0,∴3 x+
2
3 x ≥ 2
2
∴k< 2
2 -1 .