There exists an infinite set of triangles with the following properties:

(a) the lengths of the sides are integers with no common factors, and

(b) one and only one angle is $60^\circ$.

One such triangle has side lengths $5,7,8$. Find two more.
To find two more triangles with integer side lengths, no common factors, and one angle of $60^\circ$, we need to solve the equation $x^2 + 3y^2 = z^2$ for integer solutions $(x, y, z)$. The given triangle with side lengths $5, 7, 8$ corresponds to the solution $(x, y, z) = (13, 3, 14)$.

1. **Verify the given triangle:**
   - For the triangle with sides $5, 7, 8$, we have:
     \[
     5 = \frac{13 - 3}{2}, \quad 8 = \frac{13 + 3}{2}, \quad 7 = \frac{14}{2}
     \]
   - Check if $13^2 + 3 \cdot 3^2 = 14^2$:
     \[
     13^2 + 3 \cdot 3^2 = 169 + 27 = 196 = 14^2
     \]
   - This confirms that $(13, 3, 14)$ is a valid solution.

2. **Find another solution:**
   - We need to find another set of integers $(x, y, z)$ such that $x^2 + 3y^2 = z^2$.
   - Consider the solution $(x, y, z) = (37, 5, 38)$:
     \[
     37^2 + 3 \cdot 5^2 = 1369 + 75 = 1444 = 38^2
     \]
   - This is a valid solution. Now, find the corresponding triangle sides:
     \[
     a = \frac{37 - 5}{2} = 16, \quad b = \frac{37 + 5}{2} = 21, \quad c = \frac{38}{2} = 19
     \]
   - Thus, the triangle has sides $16, 19, 21$.

3. **Find yet another solution:**
   - Consider the solution $(x, y, z) = (18481, 111, 18482)$:
     \[
     18481^2 + 3 \cdot 111^2 = 341,072,161 + 36,963 = 341,109,124 = 18482^2
     \]
   - This is a valid solution. Now, find the corresponding triangle sides:
     \[
     a = \frac{18481 - 111}{2} = 9185, \quad b = \frac{18481 + 111}{2} = 9246, \quad c = \frac{18482}{2} = 9241
     \]
   - Thus, the triangle has sides $9185, 9241, 9296$.

The final answer is $\boxed{(16, 19, 21) \text{ and } (9185, 9241, 9296)}$