Answer
See the detailed answer below.
Work Step by Step
a) We can assume that the axle of rotation is frictionless, so energy is conserved.
$$E_i=E_f$$
where $E_i$ is for the hoop when its center of mass is just above the axle of rotation, while $E_f$ is when its center of mass is below the axle exactly.
$$K_i+U_{ig}=K_f+U_{gf}$$
$$\frac{1}{2}I\omega^2_i+mgy_i=\frac{1}{2}I\omega^2_f+mgy_f$$
The hoop was initially at rest;
$$\frac{1}{2}I(0^2)+mgy_i=\frac{1}{2}I\omega^2_f+mgy_f$$
the hoop's center of mass moved down a distance $2R$, thus $y_i=2R$ while $y_f=0$.
$$ 2mgR=\frac{1}{2}I\omega^2_f \tag 1$$
Now we need to find the moment of inertia of the hoop around its edge.
According to the parallel-axis theorem;
$$I=I_{cm}+mR^2= mR^2+mR^2=2mR^2$$
Plugging into (1);
$$ 2\color{red}{\bf\not} mg\color{red}{\bf\not} R=\frac{1}{\color{red}{\bf\not} 2}(\color{red}{\bf\not} 2\color{red}{\bf\not} mR^{\color{red}{\bf\not} 2})\omega^2_f $$
Thus,
$$\boxed{\omega_f=\sqrt{\dfrac{2g}{R}}}$$
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b) Hence, the speed of the lowest point is given by
$$v_T=\omega_f R_T$$
where the lowest point is moving in a vertical circle of diameter of $4R$. Hence, $R_T=2R$
$$v_T=\omega_f (2R)$$
Plugging from the boxed formula above;
$$v_T=\sqrt{\dfrac{2g}{R}}(2R)=\sqrt{\dfrac{2g(4R^{\color{red}{\bf\not} 2})}{\color{red}{\bf\not} R}}$$
$$\boxed{v_T=\sqrt{8gR}}$$