Five. (20 points) Let the functions $f(x)$ and $g(x)$ be defined as
$$
\begin{array}{l}
f(x)=12^{x}, g(x)=2000^{x} . \\
a_{1}=3, a_{n+1}=f\left(a_{n}\right)(n \in \mathbf{N}), \\
b_{1}=2000, b_{n+1}=g\left(b_{n}\right)(n \in \mathbf{N}) .
\end{array}
$$

Find the smallest positive integer $m$ such that $b_{m}>a_{2000}$.
Obviously, $12^{3}8 b_{n} \text {. }$

When $n=1$, $a_{3}=12^{16^{3}}>8 b_{1}$.
Assume that equation (1) holds for $n=k$, i.e., $a_{k+2}>8 b_{k}$.
When $n=k+1$,
$$
\begin{array}{l}
a_{k+3}=12^{a_{k+2}}>12^{8 b_{k}}=\left(12^{8}\right)^{b_{k}} \\
>(8 \times 2000)^{b_{k}}>8 \times 2000^{b_{k}}=8 b_{k+1} .
\end{array}
$$

Therefore, equation (1) holds for all $n$.
From equation (1), we get $b_{1998} \leqslant 8 b_{1998}1998 \text {. }$

In conclusion, $m=1999$.