（1）由(x-m)^2+6x=4m-3,得x2+(6-2m)x+m^2-4m+3=0.∴△=b^2-4ac=(6-2m)^2-4×1×(m^2-4m+3)=-8m+24.∵方程有实数根,∴-8m+24≥0．解得 m≤3． ∴m的取值范围是m≤3． （2）∵方程的两实根分别为x1与x2,由根与系数的关系,得 ∴x1+x2=2m-6,x1x2=m^2-4m+3,∴x1x2-(x1)^2-(x2)^2=3x1x2-(x1+x2)^2 =3x(m2-4m+3)-(2m-6)^2 =-m2+12m-27 =-(m-6)^2+9