(1)∵coA=2/3
∴sinA=√5/3
又sinB=sin[π-(A+C)]=sin(A+C)
∴sin(A+C)=√5cosC
即sinA*cosC+cosA*sinC=√5cosC
∴sinA+cosA*tanC=√5
∴tanC=(√5-sinA)/cosA
=(√5-√5/3)÷(2/3)
=√5
(2)由(1),可知tanC=√5
又sinB=√5cosC=tanC*cosC=sinC
则B=C,b=c
∴cos2B=-cosA=-2/3
即1-2sin²B=-2/3
∴sinB=±√30/6(负值舍去)
由正弦定理,有
a/sinA=b/sinB
∴b=a*sinB/sinA
=2×(√30/6)÷(√5/3)
=√6
△ABC面积S=(1/2)*b²*sinA
=(1/2)×(√6)²×(√5/3)
=√5