$\left(1\right)\because$一元二次方程${x^2}-x+\frac{1}{4}m=0$有两个实数根，
$\therefore \Delta =1-4×\frac{1}{4}m=1-m\geqslant 0$，
$\therefore m\leqslant 1$.
$\because m$为正整数，
$\therefore m=1$，
当$m=1$时，此方程为${x^2}-x+\frac{1}{4}=0$，
$\therefore$此方程的根为${x_1}={x_2}=\frac{1}{2}$.

$(2)\because$此方程的两个实数根为$a$、$b$，
$\therefore ab=\frac{1}{4}m$，${b^2}-b+\frac{1}{4}m=0$.
$\therefore y=ab-2b^{2}+2b+1=ab-2(b^{2}-b)+1=\frac{1}{4}m-2(-\frac{1}{4}m)+1=\frac{3}{4}m+1$.
解法一：$\because m=\frac{4}{3}(y-1)$，
又$\because m\leqslant 1$，
$\therefore m=\frac{4}{3}(y-1)\leqslant 1$，
$\therefore y$的取值范围为$y\leqslant \frac{7}{4}$.
解法二：
$\because m\leqslant 1$，
$\therefore \frac{3}{4}m\leqslant \frac{3}{4}$，
$\therefore \frac{3}{4}m+1\leqslant \frac{7}{4}$，
$\therefore y$的取值范围为$y\leqslant \frac{7}{4}$.

$\left(1\right)\because$一元二次方程${x^2}-x+\frac{1}{4}m=0$有两个实数根，
$\therefore \Delta =1-4×\frac{1}{4}m=1-m\geqslant 0$，
$\therefore m\leqslant 1$.
$\because m$为正整数，
$\therefore m=1$，
当$m=1$时，此方程为${x^2}-x+\frac{1}{4}=0$，
$\therefore$此方程的根为${x_1}={x_2}=\frac{1}{2}$.

$(2)\because$此方程的两个实数根为$a$、$b$，
$\therefore ab=\frac{1}{4}m$，${b^2}-b+\frac{1}{4}m=0$.
$\therefore y=ab-2b^{2}+2b+1=ab-2(b^{2}-b)+1=\frac{1}{4}m-2(-\frac{1}{4}m)+1=\frac{3}{4}m+1$.
解法一：$\because m=\frac{4}{3}(y-1)$，
又$\because m\leqslant 1$，
$\therefore m=\frac{4}{3}(y-1)\leqslant 1$，
$\therefore y$的取值范围为$y\leqslant \frac{7}{4}$.
解法二：
$\because m\leqslant 1$，
$\therefore \frac{3}{4}m\leqslant \frac{3}{4}$，
$\therefore \frac{3}{4}m+1\leqslant \frac{7}{4}$，
$\therefore y$的取值范围为$y\leqslant \frac{7}{4}$.