Let the sequence $\{x_n\}$ be defined by $x_1 \in \{5, 7\}$ and, for $k \ge 1, x_{k+1} \in \{5^{x_k} , 7^{x_k} \}$. For example, the possible values of $x_3$ are $5^{5^5}, 5^{5^7}, 5^{7^5}, 5^{7^7}, 7^{5^5}, 7^{5^7}, 7^{7^5}$, and $7^{7^7}$. Determine the sum of all possible values for the last two digits of $x_{2012}$.
1. **Initial Setup:**
   - The sequence $\{x_n\}$ is defined such that $x_1 \in \{5, 7\}$.
   - For $k \ge 1$, $x_{k+1} \in \{5^{x_k}, 7^{x_k}\}$.
   - We need to determine the sum of all possible values for the last two digits of $x_{2012}$.

2. **Case 1: $x_{2012} = 5^{x_{2011}}$**
   - Since $x_{2011} > 1$, we know that $5^{x_{2011}}$ will end in $25$ for any $x_{2011} \geq 2$.
   - Therefore, $x_{2012} \equiv 25 \pmod{100}$.

3. **Case 2: $x_{2012} = 7^{5^{x_{2010}}}$**
   - The order of $7$ modulo $100$ is $4$, meaning $7^4 \equiv 1 \pmod{100}$.
   - We need to check $5^{x_{2010}}$ modulo $4$. Since $5 \equiv 1 \pmod{4}$, we have $5^{x_{2010}} \equiv 1^{x_{2010}} \equiv 1 \pmod{4}$.
   - Thus, $x_{2012} = 7^{5^{x_{2010}}} \equiv 7^1 \equiv 7 \pmod{100}$.

4. **Case 3: $x_{2012} = 7^{7^{x_{2010}}}$**
   - Again, the order of $7$ modulo $100$ is $4$, so we need to check $7^{x_{2010}}$ modulo $4$.
   - Since $7 \equiv -1 \pmod{4}$, we have $7^{x_{2010}} \equiv (-1)^{x_{2010}} \equiv -1 \equiv 3 \pmod{4}$ (since $x_{2010}$ is always odd).
   - Therefore, $x_{2012} = 7^{7^{x_{2010}}} \equiv 7^3 \equiv 343 \equiv 43 \pmod{100}$.

5. **Summing the Possible Values:**
   - The possible values for the last two digits of $x_{2012}$ are $25$, $7$, and $43$.
   - Summing these values: $25 + 7 + 43 = 75$.

Thus, the sum of all possible values for the last two digits of $x_{2012}$ is $\boxed{75}$.