1）2f(x)≤f(6x-3)
2log2(x+1)≤log2(6x-3+1)
得: (x+1)^2≤(6x-2)
x^2+2x+1-6x+2≤0
x^2-4x+3≤0
(x-1)(x-3)≤0
1≤x≤3
经检验它满足不等式
则D={x|1≤x≤3}
2)F(x)=[log2(x)]^2-log2(x^2)=[log2(x)]^2-2log2(x)=[log2(x)-1]^2-1
当1≤x≤3时,0≤log2(x)≤log2(3),
当log2(x)=1时,即x=2时,F(x)取最小值-1
当log2(x)=0时,即x=1时,F(x)取最大值0
所以F(x)的值域为[-1,0]