|2a-b|^2=4|a|^2+|b|^2-4a·b
=4+|b|^2-4|b|cos(π/3)
=4+|b|^2-2|b|=12
即：|b|^2-2|b|-8=0
即：|b|=4
b·(2a-b)=2a·b-|b|^2=2|a|*|b|*cos(π/3)-16
=4-16=-12
b·(2a-b)=|b|*|2a-b|*cos
故：cos=b·(2a-b)/(|b|*|2a-b|)
=-12/(8√3)=-√3/2
故：=5π/6