Answer
See the detailed answer below.
Work Step by Step
a) We can find the speed by using the formula of
$$v=\sqrt{\dfrac{T_s}{\mu}}$$
where $\mu=m/L$ where $m$ is the mass of the rope and $L$ is the rope's length. Thus,
$$v=\sqrt{\dfrac{T_sL}{M}}\tag 1$$
Now we need to find $T_s$ which is seen in the figures below.
$$\sum F_y=T_s-mg=ma_y=m(0)=0$$
$$T_s=mg$$
where $m$ is the mass of the part of the rope that is under the point.
Hence,
$$T_s=\dfrac{yM}{L}g$$
Plugging into (1);
$$v=\sqrt{\dfrac{yg\color{red}{\bf\not} M\color{red}{\bf\not} L}{\color{red}{\bf\not} M\color{red}{\bf\not} L}}=\sqrt{yg}$$
$$\boxed{v=\sqrt{yg}}$$
---
b) From the previous result above, where $v=dy/dt$
$$\dfrac{dy}{dt}=\sqrt{yg}$$
Hence,
$$dt=\dfrac{dy}{\sqrt{yg}}$$
Integrating both sides:
$$\int_0^{\Delta t}dt=\int_0^L\dfrac{dy}{\sqrt{yg}}$$
$$ \Delta t =\frac{1}{\sqrt{g}}\int_0^Ly^{-\frac{1}{2}}{dy}=\frac{2}{\sqrt{g}}y^{\frac{1}{2}}\bigg|_0^L$$
$$ \Delta t =2\sqrt{L/g}$$
Thus,
$$\boxed{\Delta t=2\sqrt{\dfrac{L}{g}}}$$
