a) If $m=1$, then $k=2009^-1-n$ is solution. b) Let $1<n<N,m>1$. Then number of solutions $n<N, m>1$ is not more than $O(N^{1/2 +\epsilon})$. Therefore, the numbers n, suth that this equation had more than one solution is not more $O(N^{1/2+\epsilon})$

a) $m = 1$ and $k = 2009^n - n -1$ is solution b) Consider $n = 2009^q$ then we have $n^2|2009^n$ Suppose that $(m,k)$ is a solution of $k + m^k + n^{m^k} = 2009^n $ If $m >1$ then we get $n^2$ divides $k + m^k$, therefore $k + m^k \geq n^2 > 2n$ and $2m^k > k + m^k > 2n \Rightarrow m^k > n \Rightarrow n^{m^k} > n^n > 2009^n$, impossible So $m =1$ and $k = 2009^n - n -1$ if n is a power of 2009