Define a sequence $(x_{n})_{n\geq 1}$ by taking $x_{1}\in\left\{5,7\right\}$; when $k\ge 1$, $x_{k+1}\in\left\{5^{x_{k}},7^{x_{k}}\right\}$. Determine all possible last two digits of $x_{2009}$.
1. **Initial Condition and Sequence Definition:**
   We start with \( x_1 \in \{5, 7\} \). For \( k \geq 1 \), the sequence is defined such that \( x_{k+1} \in \{5^{x_k}, 7^{x_k}\} \).

2. **Case Analysis for \( x_{2009} = 5^{x_{2008}} \):**
   - If \( x_{2009} = 5^{x_{2008}} \), we need to determine the last two digits of \( 5^{x_{2008}} \).
   - Notice that for any positive integer \( n \geq 2 \), \( 5^n \) ends in 25. Therefore, if \( x_{2009} = 5^{x_{2008}} \), the last two digits of \( x_{2009} \) are 25.

3. **Case Analysis for \( x_{2009} = 7^{x_{2008}} \):**
   - If \( x_{2009} = 7^{x_{2008}} \), we need to determine the last two digits of \( 7^{x_{2008}} \).
   - To do this, we use the fact that \( \text{ord}_{100}(7) = 4 \). This means that \( 7^4 \equiv 1 \pmod{100} \). Therefore, the last two digits of \( 7^n \) depend on \( n \mod 4 \).

4. **Determining \( x_{2008} \mod 4 \):**
   - If \( x_{2008} = 5^{x_{2007}} \), since \( 5 \equiv 1 \pmod{4} \), we have \( 5^{x_{2007}} \equiv 1^{x_{2007}} \equiv 1 \pmod{4} \). Thus, \( x_{2008} \equiv 1 \pmod{4} \).
   - If \( x_{2008} = 7^{x_{2007}} \), we need to consider \( x_{2007} \mod 4 \):
     - If \( x_{2007} = 5^{x_{2006}} \), then \( x_{2007} \equiv 1 \pmod{4} \).
     - If \( x_{2007} = 7^{x_{2006}} \), then \( x_{2007} \equiv 3 \pmod{4} \).

5. **Possible Values for \( x_{2008} \mod 4 \):**
   - If \( x_{2008} = 7^{x_{2007}} \) and \( x_{2007} \equiv 1 \pmod{4} \), then \( 7^1 \equiv 3 \pmod{4} \).
   - If \( x_{2008} = 7^{x_{2007}} \) and \( x_{2007} \equiv 3 \pmod{4} \), then \( 7^3 \equiv 3 \pmod{4} \).

6. **Last Two Digits of \( 7^{x_{2008}} \):**
   - If \( x_{2008} \equiv 1 \pmod{4} \), then \( 7^{x_{2008}} \equiv 7^1 \equiv 7 \pmod{100} \).
   - If \( x_{2008} \equiv 3 \pmod{4} \), then \( 7^{x_{2008}} \equiv 7^3 \equiv 343 \equiv 43 \pmod{100} \).

7. **Conclusion:**
   - The possible last two digits of \( x_{2009} \) are 25, 07, and 43.

The final answer is \( \boxed{25, 07, 43} \).