5. Let the sequence $\left\{x_{n}\right\}$ satisfy $x_{1} \in\{5,7\}$ and for $k \geqslant 1$, $x_{k+1} \in\left\{5^{x_{k}}, 7^{x_{k}}\right\}$. Determine all possible values of the last two digits of $x_{2009}$.
（Liang Yingde provided）
5. Let $n=2009$. Then
(1) If $x_{n}=7^{5^{x_{n}-2}}$, then $x_{n} \equiv 7(\bmod 100)$.
In fact, since $5^{x_{n-2}} \equiv 1(\bmod 4)$, so,
$$
\begin{array}{l}
5^{x_{n-2}}=4 k+1\left(k \in \mathbf{N}_{+}\right) \text {. } \\
\text { Hence } x_{n}=7^{5^{x_{n-2}}}=7^{4 k+1}=(50-1)^{2 k} \times 7 \\
\equiv 7(\bmod 100) \text {. } \\
\end{array}
$$
(2) If $x_{n}=7^{7_{n}-2}$, then $x_{n} \equiv 43(\bmod 100)$.
In fact, since 5 and 7 are both odd numbers, we know that $x_{k}$ $(1 \leqslant k \leqslant n)$ are all odd numbers, so,
$$
7^{x_{n-2}} \equiv(-1)^{x_{n-2}} \equiv-1 \equiv 3(\bmod 4),
$$

i.e., $7^{x_{n-2}}=4 k+3\left(k \in \mathbf{N}_{+}\right)$.
Hence $x_{n}=7^{7_{n-2}^{x_{n}}}=7^{4 k+3}=(50-1)^{2 k} \times 7^{3}$
$$
\equiv 7^{3}=43(\bmod 100) \text {. }
$$
(3) If $x_{n}=5^{x_{n-1}}$, then $x_{n} \equiv 25(\bmod 100)$.
By mathematical induction, when $n \geqslant 2$,
$$
5^{n} \equiv 25(\bmod 100) \text {. }
$$

Clearly, $x_{n-1}>2$.
Therefore, $x_{n}=5^{x_{n-1}} \equiv 25(\bmod 100)$.
Combining the above results, we know that the last two digits of $x_{2000}$ can be $07, 25, 43$.