16. (13 points) A

schematic diagram of a computer device is shown in Figure 4, where $J_{1}$ and $J_{2}$ represent data inputs, and $C$ is the output for the computation result. The computation process involves inputting natural numbers $m$ and $n$ through $J_{1}$ and $J_{2}$ respectively, and after computation, a natural number $k$ is output through $C$. If this device satisfies the following three properties:
(1) When 1 is input through $J_{1}$ and $J_{2}$, the output result is 1;
(2) If a fixed natural number is input through $J_{1}$ and the natural number input through $J_{2}$ increases by 1, the output result increases by 2;
(3) If 1 is input through $J_{2}$ and the natural number input through $J_{1}$ increases by 1, the output result is doubled.

Questions: (1) If 1 is input through $J_{1}$ and a natural number $n$ is input through $J_{2}$, what is the output result?
(2) If 1 is input through $J_{2}$ and a natural number $m$ is input through $J_{1}$, what is the output result?
(3) If 2002 is input through $J_{1}$ and 9 is input through $J_{2}$, what is the output result?
16. When $J_{1}$ inputs $m, J_{2}$ inputs $n$, the output result is denoted as $f(m, n)$. Let $f(m, n)=k$, then $f(1,1)=1, f(m, n+1)$ $=f(m, n)+2, f(m+1,1)=2 f(m, 1)$.
(1) Since $f(1, n+1)=f(1, n)+2$, the sequence $f(1,1)$, $f(1,2), \cdots, f(1, n), \cdots$ forms an arithmetic sequence with the first term $f(1,1)$ and a common difference of 2.
Therefore, $f(1, n)=f(1,1)+2(n-1)=2 n-1$.
(2) Since $f(m+1,1)=2 f(m, 1)$, the sequence $f(1,1)$, $f(2,1), \cdots, f(m, 1), \cdots$ forms a geometric sequence with the first term $f(1,1)$ and a common ratio of 2.
Therefore, $f(m, 1)=f(1,1) \cdot 2^{m-1}=2^{m-1}$.
(3) Since $f(m, n+1)=f(m, n)+2$, the sequence $f(m, 1)$, $f(m, 2), \cdots, f(m, n), \cdots$ forms an arithmetic sequence with the first term $f(m, 1)$ and a common difference of 2.

Therefore, $f(m, n)=f(m, 1)+2(n-1)=2^{m-1}+2 n$ $-2, f(2002,9)=2^{2001}+16$