见解析
(Ⅰ)若a=3,b=-9,
则f'(x)=3x2-2ax+b=3x2-6x-9=3(x+1)(x-3).
令f/(x)>0,即3(x+1)(x-3)>0.则x<-1或x>3.
∴f(x)的单调增区间是(-∞,-1),(3,+∞).
令f/(x)<0,即3(x+1)(x-3)<0.则-1<x<3.
∴f(x)的单调减区间是(-1,3).
(Ⅱ)f'(x)=3x2-2ax+b,设切点为P(x,y),
则曲线y=f(x)在点P处的切线的斜率k=f'(x)=3x2-2ax+b.
由题意,知f'(x)=3x2-2ax+b=0有解,
∴△=4a2-12b≥0???a2≥3b.