解
∵sina∈[-1,1],cosa∈[-1,1]
∴sina+cosa∈[-2,2]
∴sina+cosa≠-3
∵(tana-3)(sina+cosa+3)=0
∴tana-3=0,或sina+cosa+3=0
∴tana=3或sina+cosa=-3（舍去）
2/3sin²a+1/4cos²a
=(2/3sin²a+1/4cos²a)/(sin²a+cos²a)——除以sin²a+cos²a=1,值不变
=(2/3tan²a+1/4)/(tan²a+1)——分子分母同时除以cos²a
=(2/3×9+1/4)/(9+1)
=(6+1/4)/10
=25/4×1/10
=5/8