Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the classical probability density is given by
$$
P_{\text{class}}(x) = \frac{2}{T v_{(x)} }\tag 1
$$
Where $ T $ is the period of the oscillator, and $ v_{(x)} $ is the velocity of the oscillator.
For a simple harmonic oscillator, the period $ T $ is given by
$$
T = \frac{2 \pi}{\omega}\tag 2
$$
Substitute (2) into (1);
$$
P_{\text{class}}(x) = \frac{\omega}{\pi \;v_{(x)} }\tag 3
$$
The velocity of a harmonic oscillator can be derived from the conservation of energy.
Recalling that the total mechanical energy $ E $ of a harmonic oscillator is given by
$$
E = \frac{1}{2} k A^2 = \frac{1}{2} m v^2 + \frac{1}{2} k x^2
$$
Solving for $v_{(x)} $,
$$
v_{(x)} =\sqrt{\dfrac{k (A^2 - x^2)}{m}}
$$
where $\omega = \sqrt{k/m}$, so $k=m\omega^2$
$$
v_{(x)} =\sqrt{\dfrac{m\omega^2 (A^2 - x^2)}{m}}
$$
$$
v_{(x)} =\omega\sqrt{ (A^2 - x^2)}
$$
Substituting into (3)
$$
P_{\text{class}}(x) = \frac{\omega}{\pi \omega \sqrt{A^2 - x^2}}
$$
$$
\boxed{P_{\text{class}}(x) = \frac{1}{\pi \sqrt{A^2 - x^2}}}
$$
$$\color{blue}{\bf [b]}$$
See the graph below.
$$\color{blue}{\bf [c]}$$
The probability density is highest near the turning points $ x = \pm A $, where the oscillator spends more time due to its slower velocity. Conversely, the probability density is lowest near the center $ x = 0 $, where the oscillator moves the fastest and spends the least amount of time.
