smc

First, we divide $x^8$ by $x+\frac{1}{2}$ using synthetic division or some other method. The quotient is $x^7-\frac{1}{2}{x}^6+\frac{1}{4}{x}^5-\frac{1}{8}{x}^4+\frac{1}{16}{x}^3-\frac{1}{32}{x}^2+\frac{1}{64}x-\frac{1}{128}$, and the remainder is $\frac{1}{256}$. Then we plug the solution to $x+\frac{1}{2}=0$ into the quotient to find the remainder. Notice that every term in the quotient, when $x=-\frac{1}{2}$, evaluates to $-\frac{1}{128}$. Thus $r_2=-\frac{8}{128}=\boxed{-\frac{1}{16}}$.