解:
(1)
由tanA+tanC+tan(π/3)=tanAtanCtan(π/3)
可以得出 tanA+tanC=-√3*(1-tanAtanC)
(tanA+tanC)/(1-tanAtanC)=tan(A+C)=-√3
在三角形中 tanB=-tan(A+C)=√3
∴B=π/3
(2)
正弦定理a/sinA=b/sinB=c/sinC
∴(a+c)/(sinA+sinC)=b/sinB=(√3/2)/(√3/2)=1
即a+c=sinA+sinC
∵B=π/3
∴C=2π/3-A
a+c=sinA+sin(2π/3-A) 展开再化简
a+c=√3sin(A+π/6)
∵A∈(0,2π/3)
∴A+π/6∈(π/6,5π/6)
∵1/2<3sin(A+π/6)≤1
∴a+c的取值范围是:(√3/2,√3]