Rope on a table .....

Since the rope is uniform, the mass of the rope is proportional to it's length. Using the friction equation, we know that :

$f = N \cdot \mu = m_{1}g \cdot \mu$

The rope begins to slide when the mass of the hanging rope is 25% the total mass of the rope thus :

$f = w$

$m_{1}g \cdot \mu = m_{2}g$

$\frac{3}{4}Mg \cdot \mu = \frac{1}{4}Mg$

$\frac{3}{4} \cdot \mu = \frac{1}{4}$

$\mu = \frac{1}{3} = \boxed{0.33}$

Details and assumptions : $m_{1}$ is the mass of the rope that is on the table; $m_{2}$ is the mass of the rope that is hanging on the side; $M$ is the total mass of the rope