(1)f(x)=[f(x/2)]^2>=0;f(0+x)=f(0)f(x),所以f(0)=1;而f(0)=f(-x+x)=f(-x)*f(x);f(-x)=1/f(x);所以不管x>0,或x<0,或x=0,都有f(x)>0(2)f(x)=f(y)*f(x-y),若x>y,则0(3)A={x|f(x^2-6x+8)>1}={x|x^2-6x+8<0}=(2,4)B={x|f(x^2-4ax+3a^2)>1}=(3a,a),a<0 (a,3a),a>0而A交B=0,所以a<0或0=4
