Five. (20 points) Given the sequence of positive integers $\left\{T_{m, n}\right\}$ satisfies:
(1) $T_{m, m}=2^{m}$;
(2) $T_{m, n}=T_{n, m}$;
(3) $\left(T_{m, n}\right)^{m+n}=\left(T_{m, m+n}\right)^{n}$.

Here $m, n$ are any positive integers. Try to find the general term formula for the sequence $\left\{T_{m, n}\right\}$.
Five, $[m, n]$ is the least common multiple of $m$ and $n$.
Assume $n \geqslant m$, let $n=q m+r, 0 \leqslant r<m$. From (3) we know $\left(T_{m, q n+r}\right)^{(\varphi+1) m+r}=\left(T_{m,(q+1) m+r}\right)^{m m+r}$,
i.e., $\left(T_{m \cdot q m+r}\right)^{\frac{1}{m+r}}=\left(T_{m \cdot(q+1) m+1)^{(4+1) m+r}}\right.$.
Let $S_{q}=\left(T_{m, q m+r}\right)^{\frac{1}{m+1}}$, then $S_{q+1}=S_{q}$.
Thus, $S_{q}=S_{0}=\left(T_{m, r}\right)^{\frac{1}{r}}$. Therefore, we have
$\left(T_{m, m+r}\right)^{\frac{1}{m+r}}=\left(T_{m, a t}\right)^{\frac{1}{4}}=\left(T_{m, r}\right)^{\frac{1}{r}}$.
Let $m=q_{1} r+r_{1}$.
Similarly, $\left(T_{r, m}\right)^{\frac{1}{m}}=\left(T_{r, r_{1}}\right)^{\frac{1}{r_{1}}}$.
From (2) we know
$T_{m, r}=T_{r, m}$, hence $T_{m, n}=\left(T_{r, r_{1}}\right)^{\frac{m}{\pi_{1}}}$.
Sequentially set $r=q_{2} r_{1}+r_{2}, r_{1}=q_{3} r_{2}+r_{3}, \cdots, r_{k-1}=$ $q_{k+1} r_{k}$, where $r_{k}=(m, n)$ is the greatest common divisor of $m$ and $n$.
Recursively, we have
$$
\begin{array}{l}
T_{m, n}=\left(T_{r_{1}, .}\right)^{\frac{m}{m_{1}}}=\left(T_{r_{1} \jmath_{2}}\right)^{\frac{\operatorname{mom}}{\pi_{1} r_{2}}}=\cdots \\
=\left(T_{r_{k-1} r_{k}}\right)^{\frac{m a v \cdots r_{k-2}}{\pi_{1} \cdots r_{k}}}=\left(T_{r_{k-1} r_{k}}\right)^{\frac{m m}{r_{k-1} r_{k}}} \\
=\left(T_{r_{k}-r_{k}}\right)^{\frac{m m-1}{r_{k}-r_{k}} \cdot \frac{r_{k-1}}{r_{k}}}=2^{\frac{\operatorname{mn}}{r_{k}}}=2^{\frac{m m}{(m \cdot m)}}=2^{(m \cdot n)} . \\
\end{array}
$$

Next, verify that it satisfies the conditions in the problem. Conditions (1) and (2) are obviously satisfied. For condition (3), i.e.,
$$
\begin{array}{l}
2^{(m+n)[m \cdot n)}=2^{n!m, m+n)} \\
\Leftrightarrow(m+n)[m, n]=n[m, m+n] .
\end{array}
$$

Using $[m, n]=\frac{m n}{(m, n)}$, we know it is equivalent to $(m, n)=$ $(m, m+n)$ being true.