$解：?(1)?∵方程?x2-2x+m=0?有兩個實數(shù)根$
$∴?△=(-2)2-4m≥0?，解得?m≤1?$
$故?m ?的取值范圍是?m≤1?$
$?(2)?由根與系數(shù)的關系可知$
$?x_{1}+x_{2}=2?，?x_{1}x_{2}=m?$
$∵?x_{1}+3x_{2}=3?$
$∴?2+2x_{2}=3?，解得?x_{2}=\frac {1}{2}?$
$∴?x_{1}=2-\frac {1}{2}=\frac {3}{2}?$
$∴?m=x_{1}x_{2}=\frac {3}{4}?$
