(1) f(x)在[0,+∞)关联,在[0,1]不关联,任取x2-x1∈[0,+∞),则f(x2 )-f(x1 )=2(x2-x1 )∈[0,+∞),∴f(x)在[0,+∞)关联.取x1=1,x2=0,则x1-x2∈[0,1],而f(x1 )-f(x2 )=2(x1-x2 )=2∉[0,1],∴f(x)在[0,1不关联.(2)∵f(x)在{3}关联,∴对于任意x1-x2=3,都有f(x1 )-f(x2 )=3,∴对任意x,都有f(x+3)-f(x)=3.由x∈[0,3)时, f(x)=x^2-2x,得f(x)在x∈[0,3)的值域为[-1,3),∴f(x)在x∈[3,6)的值域为[2,6),∴2≤f(x)≤3仅在x∈[0,3)或x∈[3,6)上有解,x∈[0,3)时, f(x)=x^2-2x,令2≤x^2-2x≤3,解得√3+1≤x<3,x∈[3,6)时, f(x)=f(x-3)+3=x^2-8x+18,令2≤x^2-8x+18≤3,解得3≤x≤5，∴不等式2≤f(x)≤3的解为[√3+1,5].(3)证明：①先证明：f(x)是在{1}关联的,且是在[0,+∞)关联的→f(x)在[1,2]是关联的，∵f(x)是在{1}关联的，∴当x1-x2=1时, f(x1 )-f(x2 )=1,即f(x+1)-f(x)=1,∵f(x)是在[0,+∞)关联的，∴当x1-x2≥0时, f(x1 )-f(x2 )≥0,任取x1-x2∈[1,2],即1≤x1-x2≤2，∴x1≥x2+1,x1≤x2+2,∴f(x2+1)≤f(x1 )≤f(x2+2),∴f(x1 )-f(x2 )≥f(x2+1...