（1）由向量m⊥向量n可得：
-2cos(A+π/6)-sin(A-π/3)=0,
∵cos(A+π/6)=cos[π/2+(A-π/3)]=-sin(A-π/3)
∴sin(A-π/3)=0,
∵A∈(0,π),∴A-π/3=0,即A=π/3.
（2）(sin²C-cos²C)/(1-sin2C)=-2
即(sin²C-cos²C)/(sinc-cosc)²=-2,
∴(sinC+cosC)/(sinC-cosC)=-2,
∴(tanC+1)/(tanC-1)=-2,
解得：tanC=1/3.
tanB=tan[π-(A+C)]=-tan(A+C)=-(tanA+tanc)/(1-tanAtanC)=-(6+5√3)/3