Answer
a) $21.8\;\rm rad/s^2$
b) $6.6\;\rm rad/s$
Work Step by Step
a) The only force exerted on the disk at this moment is its own eight.
Thus,
$$\sum \tau=mgR=I\alpha$$
We chose clockwise to be the positive direction.
Solving for $\alpha$;
$$\alpha=\dfrac{mgR}{I}\tag 1$$
Now we need to find the moment of inertia of the disk at its end as seen in the figure below.
According to the parallel-axis theorem;
$$I=I_{cm}+mR^2$$
$$I=\frac{1}{2}mR^2+mR^2=1.5mR^2\tag 2$$
Plugging into (1);
$$\alpha=\dfrac{\color{red}{\bf\not} mg\color{red}{\bf\not} R}{1.5\color{red}{\bf\not} mR^{\color{red}{\bf\not} 2}}=\dfrac{g}{1.5R} $$
Plugging the known;
$$\alpha=\dfrac{9.8}{1.5\times0.30 }=\color{red}{\bf21.8}\;\rm rad/s^2$$
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b) We can assume that the axle of rotation is frictionless, so energy is conserved.
$$E_i=E_f$$
where $E_i$ is for the disk when its center of mass is parallel to the axle and at the same level as it, while $E_f$ is when its center of mass is below the axle exactly.
$$K_i+U_{ig}=K_f+U_{gf}$$
$$\frac{1}{2}I\omega_i^2+mgy_i=\frac{1}{2}I\omega_f^2+mgy_f$$
The disk was initially at rest;
$$\frac{1}{2}I(0)^2+mgy_i=\frac{1}{2}I\omega_f^2+mgy_f$$
the disk's center of mass moved down a distance $R$, thus $y_i=R$ while $y_f=0$.
$$ mgR=\frac{1}{2}I\omega_f^2+mg(0)$$
Plugging from (2);
$$ \color{red}{\bf\not} mg\color{red}{\bf\not} R=\frac{1}{2}\left(1.5\color{red}{\bf\not} mR^{\color{red}{\bf\not} 2}\right)\omega_f^2 $$
Hence,
$$\omega_f=\sqrt{\dfrac{2g}{1.5R}}=\sqrt{\dfrac{2\times 9.8}{1.5\times 0.3}}$$
$$\omega_f= \color{red}{\bf 6.6}\;\rm rad/s$$
