在△ABC中,角A,B,C所对的对边长分别为a,b,c.
（1）设向量x=(sinB,sinC),向量y=(cosB,cosC),向量z=(cosB,-cosC),若z‖（x+y）,求tanB+tanc的值
（2）已知a²-c²=8b,且sinAcosC+3cosAsinC=0,求b
1,x+y=(sinB+cosB,sinC+cosC)
z||(x+y),(sinB+cosB)/cosB=-(sinC+cosC)/cosC
tanB+1=-tanC-1 tanB+tanC=-2
2,sinAcosC+3cosAsinC=sin(A+C)+2cosAsinC
=sinB+2cosAsinC
=b+(b^2+c^2-a^2)/2bc*c*2=0
b=4