（1）将x=y=0代入 f(x+y)+f(x-y)=2f(x)f(y)得,f(0)+f(0)=2f(0)f(0),即
2f(0)=2f(0)f(0),由f(0)≠0得,f(0)=1,
再将x=0代入f(0+y)+f(0-y)=2f(0)f(y)得,
f(y)+f(-y)=2f(y),f(-y)=f(y),故f(x)是偶函数.
（2）将y=1/2代入f(x+y)+f(x-y)=2f(x)f(y)得
f(x+1/2)+f(x-1/2)=2f(x)f(1/2)=0.由f(1/2)=0 得 f(x+1/2)+f(x-1/2) =0.
f(x+1/2)=-f(x-1/2) ,同理f(x-1/2)= f(x-1+1/2)=- f(x-1-1/2)= - f(x-3/2),故f(x+1/2)= -f(x-1/2) =f(x-3/2),
设y= x-3/2,则x=y+3/2,x+1/2=y+2,代入上式得f(y+2)=f(y),故f(x)是周期为2的周期函数.
（3）设x=y代入f(x+y)+f(x-y)=2f(x)f(y)得f(2x)+f(0)=2f(x)f(x),
f(2x) =2f(x)f(x)-1,将x=1/2代入得f(1) =2f(1/2)f(1/2)-1=-1,即
f(1) =-1,
设y=2x代入f(x+y)+f(x-y)=2f(x)f(y)得,f(3x)+f(-x)=2f(x)f(2x),故
f(3x)=2f(x)f(2x)-f(x)=f(x)( 2f(2x)-1)
=f(x)( 4f(x)f(x)-3)=4f(x)f(x)f(x)-3f(x)
将x=1/3代入上式得
f(1)=4 f(1/3) f(1/3) f(1/3)-3 f(1/3),设a=f(1/3)得
4a^3-3a+1=0,解得a=1/2,即f(1/3)=1/2,
再x=1/6代入f(2x) =2f(x)f(x)-1得f(1/3) =2f(1/6)f(1/6)-1,设b=f(1/3)得2bb-1= f(1/3)=1/2,4bb-3=0,b=√3/2.f(x)在[0,1]内单调t得f(1/6)=√3/2.
