jmc

Setting $x=1,$ we get $$e(1)+o(1)=\frac{6}{1+2}+1^2+2^1=5.$$Setting $x=-1,$ we get $$e(-1)+o(-1)=\frac{6}{-1+2}+(-1{)}^2+2^{-1}=\frac{15}{2}.$$Since $e(x)$ is an even function and $o(x)$ is an odd function, $e(-1)=e(1)$ and $o(-1)=-o(1),$ so $$e(1)-o(1)=\frac{15}{2}.$$Subtracting this from the equation $e(1)+o(1)=5,$ we get $$2o(1)=-\frac{5}{2},$$so $o(1)=\boxed{-\frac{5}{4}}.$