解:∵向量m∥向量n, ∴b√3*sinA-acosB=0.
√3sinBsinA-sinAcosB=0.
√3tanB-1=0.
tanB=√3/3.
∠B=30°;
若b=2. ...
由正弦定理,得:a/sinA=b/sinB=c/sinC.
a=bsinA/sinB, c=bsinC/sinB.
a+c=b(sinA+sinC)/sinB.
=4(sinA+sinC).
=4[2sin(A+C)/2*cos(A-C)/2].
=8cosB/2*cos(A-C)/2.
当cos(A-C)/2=1时,a+c取得最大值,a+c的最大值为8cosB/2.
即,(a+c)max=8cosB/2.
cosB/2=±√[(1+cosB)/2]
=√[(1+cos30°)/2]. ( 只取“+”值]
=(1/2).√(2+√3).
∴(a+c)max=4√(2+√3).