16.(1)bsinAcosB=(2c-b)sinBcosA,
由正弦定理,sinBsinAcosB=(2sinC-sinB)sinBcosA,sinB>0,
∴sinAcosB=2cosAsinC-cosAsinB,
∴sinAcosB+cosAsinB=2cosAsinC,
sinAcosB+cosAsinB=sin(A+B)=sinC>0,
∴cosA=1/2,
∴A=60°.
(2)易知|m|=|n|=1,
mn=sinBcos2C+cosBsin2C=sin(B+2C)=sin(120°+C),
0