(1)∵m·n=1.
∴-cosA+√3sinA=1
∴2sin(A-π/6)=1
∴A-π/6=π/6或5π/6
∵A∈(0,π)∴A=π/3
(2)(1+sin2b)/〔(cosB)^2-(sinB)^2〕=(sinB+cosB)^2/(sinB+cosB)(cosB-sinB)=(sinB+cosB)/(cosB-sinB)=-3
∴tanB+1=-3(1-tanB)
∴tanB=2 ∵B+C=π-π/3=2π/3
∴tanC=tan(2π/3-B)=(tan2π/3-tanB)/1+tan2π/3*tanB=(-√3-2)/1-2√3=(8+5√3)/11