(1) ∵向量m与向量n公线 ∴cosB/2a-b=cosC/c 即cosB.c-(2a-b).cosC=0
在△ABC中,由正弦定理得：a/ sinA =b /sinB =c/sinC =k≠0
∴a=ksinA,b=ksinB,c=ksinC,代入得
k[cosB.sinC-2sinAcosC+sinBcosC]=k[sin（B+C)-2sinAcosC]=ksinA(1-2cosC)=0
∵ABC均∈（0,π）,∴sinA≠0,1-2cosC=0 即cosC=1/2 解得C=π/3
(2)由（1）中可得 k=c/sinC=4
a+b=ksinA+ksinB=4[sin(2π/3 - B)+sinB]=4√3sin(π/3+B)
∵B∈（0,2π/3 ) ∴当B =π/3时 有a+b（max）=4√3