(1) f(8)=f(4)+f(2)
=f(2)+f(2)+f(2)
=3.
(2)f(x)-f(x-2)>3,
f(x)>f(x-2)+f(8),
f(x)>f[8(x-2)],
函数在(0,+∞)是增函数,
x>8(x-2),且x>0,8(x-2)>0,
三式解得
2