(1)∵函数f(x)=x3-4x2+5x-4,
∴f′(x)=3x2-8x+5,
根据导数的几何意义,则曲线f(x)在x=2处的切线的斜率为f′(2)=1,
又切点坐标为(2,-2),
由点斜式可得切线方程为y-(-2)=1×(x-2),即x-y-4=0,
∴求曲线f(x)在x=2处的切线方程为x-y-4=0;
(2)设切点坐标为P(a,a3-4a2+5a-4),
由(1)可知,f′(x)=3x2-8x+5,
则切线的斜率为f′(a)=3a2-8a+5,
由点斜式可得切线方程为y-(a3-4a2+5a-4)=(3a2-8a+5)(x-a),①
又根据已知,切线方程过点A(2,-2),
∴-2-(a3-4a2+5a-4)=(3a2-8a+5)(2-a),即a3-5a2+8a-4=0,
∴(a-1)(a2-4a+4)=0,即(a-1)(a-2)2=0,
解得a=1或a=2,
将a=1和a=2代入①可得,切线方程为y+2=0或x-y-4=0,
故经过点A(2,-2)的曲线f(x)的切线方程为y+2=0或x-y-4=0.