(1)x-y-4=0
(2)x-y-4=0或y+2=0
(1)∵f′(x)=3x 2-8x+5,
∴f′(2)=1,又f(2)=-2,
∴曲线f(x)在点(2,f(2))处的切线方程为y-(-2)=x-2,即x-y-4=0.
(2)设切点坐标为(x 0,x 0 3-4x 0 2+5x 0-4),
∵f′(x 0)=3x 0 2-8x 0+5,
∴切线方程为y-(-2)=(3x 0 2-8x 0+5)(x-2),
又切线过点(x 0,x 0 3-4x 0 2+5x 0-4),
∴x 0 3-4x 0 2+5x 0-2=(3x 0 2-8x 0+5)(x 0-2),
整理得(x 0-2) 2(x 0-1)=0,解得x 0=2或x 0=1,
∴经过A(2,-2)的曲线f(x)的切线方程为x-y-4=0或y+2=0.