设函数y=f(x)定义在R上,对于任何实数m.n,恒有f(m+n)=f(m)*f(n),且当x大于零时,0＜f(x)＜1

(1)证明：令n=0,m>0时,则f(m)=f(m)f(0),且00,令y>0,则有0＜f(y)＜1f(x+y)=f(x)*f(y)y,所以f(x)在R上是减函数(3)f(-x^2+6x-1)*f(y)=1=f(0)=f(y-y)=f(y)*f(-y)而f(x)在R上是减函数所以必有f(-x^2+6x-1)=f(-y)且-x^2+6x-1=-y整理为y=x^2-6x+1则y=x^2-6x+1=(x-3)^2-8>=-8而,B=｛(x,y)「y=a｝,A∩B＝空集所以只需a