（1）f(x)=2cos²x+2√3sinxcosx
=2cos²x+√3sin2x
=cos2x+1+√3sin2x
=2sin(2x+π/6)+1
∵x∈[-π/6,π/3].∴2x+π/6∈[-π/6,π]
∴f(x)∈[0,3]
（2）∵cos(A-C)-cos(A+C),
=cosAcosC+sinAsinC-(cosAcosC-sinAsinC)
=2sinAsinC
=2sinB
∴sinAsinC=sinB
f(C)=2sin(2C+π／6)+1=2
所以sin(2C+π／6)=1/2
即2C+π/6=π/6或5π/6
当2C+π/6=π/6,C=0
不符舍去
所以2C+π/6=5π/6,C=π/3
sinAsinC=sinB=sin(A+C)
即(√3sinA)/2=sin(A+π/3)=(sinA)/2+(√3cosA)/2
(√3-1)sinA=√3cosA
tanA=√3/(√3-1)= 3-√3