(1)f'(x)=6x^2+6ax+3b.因为函数f(x)在x=1及x=2时取得极值,故有{f'(1)=0,f'(2)=0}--->{6+6a+3b=0,24+12a+3b=0}--->a=-3,b=4(2)由以上知,f(x)=2x^3-9x^2+12x+8c,f'(x)=6x^2-18x+12=6(x-1)(x-2)当x属于[0,1)时,f'(x)>0...