解:【1】由題設(shè)可知：當(dāng)0≤x≤1時,f(x)=x(1-x)∴當(dāng)x=0.5時,有f(0.5)=0.5(1-0.5)=0.25=1/4即f(0.5)=1/4【2】由題設(shè)可知：對任意實(shí)數(shù)x,恒有f(x+1)=2f(x).∴f(x+2)=f[(x+1)+1]=2f(x+1)=4f(x)即...為什么f(x+1)=2f(x).就∴f(x+2)=f[(x+1)+1]？？？啊,你可以這樣,∵恒有f(x+1)=2f(x)∴令x=t+1.就有f[(t+1)+1]=2f(t+1)即f(t+2)=2f(t+1)又顯然有f(t+1)=2f(t).∴f(t+2)=4f(t)即f(x+2)=4f(x)