Consider a standard twelve-hour clock whose hour and minute hands move continuously. Let $m$ be an integer, with $1 \leq m \leq 720$. At precisely $m$ minutes after 12:00, the angle made by the hour hand and minute hand is exactly $1^\circ$.
Determine all possible values of $m$.
1. **Determine the angle made by the minute hand:**
   - The minute hand moves $6$ degrees per minute. Therefore, after $m$ minutes, the angle made by the minute hand is:
   \[
   \theta_m = 6m \text{ degrees}
   \]

2. **Determine the angle made by the hour hand:**
   - The hour hand moves $0.5$ degrees per minute (since it moves $30$ degrees per hour and there are $60$ minutes in an hour). Therefore, after $m$ minutes, the angle made by the hour hand is:
   \[
   \theta_h = 0.5m \text{ degrees}
   \]

3. **Set up the equation for the angle difference:**
   - The problem states that the angle between the hour and minute hands is exactly $1^\circ$. Therefore, we have:
   \[
   |6m - 0.5m| = 1 + 360k \quad \text{or} \quad |6m - 0.5m| = 1 - 360k \quad \text{for some integer } k
   \]
   Simplifying the absolute value expression, we get:
   \[
   |5.5m| = 1 + 360k \quad \text{or} \quad |5.5m| = 1 - 360k
   \]

4. **Solve the equations:**
   - We can rewrite the equations as:
   \[
   5.5m = 1 + 360k \quad \text{or} \quad 5.5m = -1 + 360k
   \]
   - Solving for $m$ in each case:
   \[
   m = \frac{1 + 360k}{5.5} \quad \text{or} \quad m = \frac{-1 + 360k}{5.5}
   \]

5. **Simplify the expressions:**
   - For the first equation:
   \[
   m = \frac{1 + 360k}{5.5} = \frac{2 + 720k}{11}
   \]
   - For the second equation:
   \[
   m = \frac{-1 + 360k}{5.5} = \frac{-2 + 720k}{11}
   \]

6. **Check for integer solutions:**
   - Since $m$ must be an integer, $2 + 720k$ and $-2 + 720k$ must be divisible by $11$.
   - For $2 + 720k$:
     \[
     2 + 720k \equiv 0 \pmod{11} \implies 720k \equiv -2 \pmod{11} \implies 5k \equiv -2 \pmod{11}
     \]
     Solving for $k$:
     \[
     5k \equiv 9 \pmod{11} \implies k \equiv 9 \cdot 5^{-1} \pmod{11} \implies k \equiv 9 \cdot 9 \pmod{11} \implies k \equiv 4 \pmod{11}
     \]
     Therefore, $k = 4 + 11n$ for some integer $n$.
   - For $-2 + 720k$:
     \[
     -2 + 720k \equiv 0 \pmod{11} \implies 720k \equiv 2 \pmod{11} \implies 5k \equiv 2 \pmod{11}
     \]
     Solving for $k$:
     \[
     5k \equiv 2 \pmod{11} \implies k \equiv 2 \cdot 5^{-1} \pmod{11} \implies k \equiv 2 \cdot 9 \pmod{11} \implies k \equiv 7 \pmod{11}
     \]
     Therefore, $k = 7 + 11n$ for some integer $n$.

7. **Calculate the corresponding values of $m$:**
   - For $k = 4$:
     \[
     m = \frac{2 + 720 \cdot 4}{11} = \frac{2 + 2880}{11} = \frac{2882}{11} = 262
     \]
   - For $k = 7$:
     \[
     m = \frac{-2 + 720 \cdot 7}{11} = \frac{-2 + 5040}{11} = \frac{5038}{11} = 458
     \]

8. **Verify the range of $m$:**
   - Both $262$ and $458$ are within the range $1 \leq m \leq 720$.

The final answer is $\boxed{262}$ and $\boxed{458}$.