解:$(1)M=(2x²+3xy+2y)-2(x²-xy+x-\frac {1}{2})$
$=2x²+ 3xy+2y-2x²+2xy-2x+1$
$=5xy+2y-2x+1.$
当$x=\frac {1}{5}$,$y=-1$时,
原式$=5×\frac {1}{5}×(-1)+2×(-1)-2×\frac {1}{5}+1= -1-2-\frac {2}{5}+1=-2\frac {2}{5}.$
$(2)$因为$M=5xy+2y-2x+1=(5y-2)x+2y+1$,多项式$M$的值与字母$x$的取值无关,
所以$5y-2=0$,
解得$y=\frac {2}{5}.$