$解:(1)当m=1时,\frac {x}{x-1}-2=\frac{1}{1-x}$
$\frac{x}{x-1}-\frac{1}{1-x}=2,\frac {x}{x-1}+\frac{1}{x-1}=2,\frac{x+1}{x-1}=2$
$去分母得:x+1=2(x-1),解得:x=3$
$检验:当x=3时,x-1≠0,故方程的解为x=3$
$(2)\frac{x}{x-1}+\frac{m}{x-1}=2,\frac {x+m}{x-1}=2$
$去分母得:x+m=2(x-1),解得:x=m+2$
$∵分式方程有解且解为非负数,x≠1且x≥0, ∴m+2≠1且m+2≥0,即m≥-2且m≠-1$