Answer
$\frac{1}{16}$, $(x-(1/4))^2$
Work Step by Step
To complete the square of $x^2 -\frac{1}{2}x$ we add $(b/2)^2$. In this case, the "$b$" is the second coefficient, $-1/2$, so we add $((-1/2)/2)^2=1/16$ and we have $x^2 -\frac{1}{2}x+\frac{1}{16}$. Then by factorization, we get
$$
x^2 -\frac{1}{2}x+\frac{1}{16}=(x-(1/4))^2
.$$