(1)f'(x)=x^2-(2a+1)x+a^2+a,令f'(x)=0,得x=a或x=a+1,由题意知f(x)在（－∞,a）、（a+1,+∞）上单调递增,在（a,a+1）上单调递减,故f(x)在x=a处取得极大值,故a=1
(2)由题意知,f'(x)=k无解,即x^2-(2a+1)x+a^2+a-k=0无解,所以(2a+1)^2-4(a^2+a-k)=4k+1＜0,得k＜-1/4
(3)当-1＜a＜0时,f(x)在[0,a+1]上单调递减,(a+1,1)上单调递增,f(0)=0,f(1)=a^2-1/6故当-1＜a≤-√6/6时,f(x)max=f(1)=a^2-1/6,当-√6/6＜a＜0时,f(x)max=f(0)=0
当0≤a＜1时,f(x)在[0,a]上单调递增,在(a,1)上单调递减,故f(x)max=f（a）=1/3a^3+1/2a^2
当a≥1时,f(x)在区间[0,1]上单调递增,故f(x)max=f(1)=a^2-1/6