jmc

Since $2^n$ is a power of $2$, its only prime factor is $2$. So every odd integer is invertible modulo $2^n$ and every even integer is non-invertible modulo $2^n$. Among the positive integers less than $2^n$, there are precisely $\frac{2^n}{2}=2^{n-1}$ odd integers. Thus, $$k=2^{n-1}\equiv 2^{-1}{2}^n\equiv 7\cdot 3\equiv 21\equiv \boxed{8}(\mathrm{mod}13)$$