∵a∥b,∴√3cosa-sina+m=0,∴m=sina-√3cosa=2sin(a-π/3)
∵a∈(0,π/2),∴a-π/3∈(-π/3,π/6),∴sin(a-π/3)∈(-√3/2,1/2)
∴m∈(-√3,1),故最小值不存在.
（2）∵a∥b,且m=0,∴sin(a-π/3)=0,∴a=π/3
∴cos(π/2-a)*sin(π+2a)/cos(π-a)
=sina*(-sin2a)/(-cosa)
=2sin²a*cosa/cosa
=2sin²a
=2×(√3/2)²
=3/2