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}} $.