1、解：因为sin^2b>=0,sin^2a>=0;所以：2sin^2b+3sin^2a>=0.即：2sina>=0,得到：sina>=0.∵2sin^2b+3sin^2a=2sina，∴sin^2b=sina-(3/2)sin^2a>=0进一步：Sina(1-3/2sina)>=01-3/2sina>=0,得到：sina<=2/3.即sina的取值范围为：[0,2/3].
2、则：m=sin^2a+sin^2b=sin^2a+sina-(3/2)sin^2a=-(1/2)sin^2a+sina=-(1/2)(sin^2a-2sina+1)+1/2=-(1/2)(sina-1)^2+1/2.因为0<=sina<=2/3.所以：当sina=2/3，m有最大值m=4/9当sina=0，m有最小值m=0。所以m的取值范围为：[0,4/9].
