(1)f'(x)=x^2-(2a+1)x+a^2+a,令f'(x)=0,得x=a或x=a+1,由題意知f(x)在（－∞,a）、（a+1,+∞）上單調(diào)遞增,在（a,a+1）上單調(diào)遞減,故f(x)在x=a處取得極大值,故a=1
(2)由題意知,f'(x)=k無(wú)解,即x^2-(2a+1)x+a^2+a-k=0無(wú)解,所以(2a+1)^2-4(a^2+a-k)=4k+1＜0,得k＜-1/4
(3)當(dāng)-1＜a＜0時(shí),f(x)在[0,a+1]上單調(diào)遞減,(a+1,1)上單調(diào)遞增,f(0)=0,f(1)=a^2-1/6故當(dāng)-1＜a≤-√6/6時(shí),f(x)max=f(1)=a^2-1/6,當(dāng)-√6/6＜a＜0時(shí),f(x)max=f(0)=0
當(dāng)0≤a＜1時(shí),f(x)在[0,a]上單調(diào)遞增,在(a,1)上單調(diào)遞減,故f(x)max=f（a）=1/3a^3+1/2a^2
當(dāng)a≥1時(shí),f(x)在區(qū)間[0,1]上單調(diào)遞增,故f(x)max=f(1)=a^2-1/6