解（1）：∵tana=√3,π＜a＜3π/2
∴a=π+π/3=4π/3
∴cosa-sina=cos(4π/3)-sin(4π/3)=-cos（π/3)+sin(π/3)=-1/2+√3/2
解（2）∵tanα=-2
∴cosα≠0
∴（3sinα-2cosα)/(5sinα+4cosα),分子、分母同时除以cosα,得：
=（3sinα/cosα-2)/(5sinα/cosα+4),∵tanα=sinα/cosα=-2
=(3tanα-2)/(5tanα+4)
=[3×(-2)-2]/[5×（-2）+4]
=（-8）/(-6)
=4/3