f(x)=√3sin^2 x+sinxcosx-(√3/2)（x∈R）
=√3*[(1-cos2x)/2]+(1/2)sin2x-(√3/2)
=(√3/2)-(√3/2)cos2x+(1/2)sin2x-(√3/2)
=(1/2)sin2x-(√3/2)cos2x
=sin[2x-(π/3)]
(1)f(π/4)=sin[(π/2)-(π/3)]=sin(π/6)=1/2
(2)当x∈(0,π)时,2x∈(0,2π)
2x-(π/3)∈(-π/3,5π/3)
所以,f(x)的最大值为1
(1)A＜B,且f(A)=f(B)=1/2
===> sin[2A-(π/3)]=sin[2B-(π/3)]=1/2
===> 2A-(π/3)=π/6；2B-(π/3)=5π/6
===> 2A=π/2；2B=7π/6
===> A=π/4,B=7π/12
所以,C=π-(π/4)-(7π/12)=π/6
在△ABC中,BC=a,AB=c
所以,由正弦定理有：BC/AB=a/c=sinA/sinC=√2.