解由tan2a=-4/3
知tan2a=2tana/[1-tan^2a]=-4/3
即3tana=2tan^2a-2
即2tan^2a-3tana-2=0
即（2tana+1）（tana-2）=0
即tana=-1/2或tana=2
由-π/2＜a＜0
知tan=-1/2
故[sin（a+π/6)]/[sin a+cos a]
=[sina×√3/2+cosa×1/2]/[sin a+cos a]
=[tana×√3/2+1/2]/[tana+1]
=[（-1/2)×√3/2+1/2]/[(-1/2)+1]
=1-√3/2