Hint: $F=m\omega^2 x$

Step: Analyze each statement one by one.
 
The displacement of the particle is zero at times $\frac{T}{4}, \frac{3T}{4}, \frac{5T}{4}$.
The acceleration of the particle performing SHM is given by;
$\Rightarrow a = -\omega^2 x$
When the displacement of the particle is zero, then the particle's acceleration is also zero i.e., at $t = \frac{T}{4}, \frac{3T}{4}, \frac{5T}{4}$, the acceleration of the particle is zero.
According to Newton's second law of motion, the force acting on the particle is given by;
$F = ma$
$\Rightarrow F = 0~\text{if}~a =0$
So, the force is zero at $t=\frac{3T}{4}.$
The displacement of the particle is maximum at times $t = 0,\frac{2T}{4}, T$
So, the acceleration of the particle is maximum at $t = T$.
The velocity of the particle is given by; 
$\Rightarrow v = \omega\sqrt{A^2-x^2}$
The velocity of the particle is maximum $(v_{max}= A\omega)$ when the displacement of the particle is zero i.e., at $t = \frac{T}{4}, \frac{3T}{4}, \frac{5T}{4}$.
The potential energy is equal to the kinetic energy of the oscillation i.e., 
$\Rightarrow \frac{1}{2} m\omega^2 x^2 = \frac{1}{2}m\omega^2(A^2- x^2)$
$\Rightarrow x = \frac{A}{\sqrt{2}}$
At $t = \frac{T}{2}$, the particle is at an extreme position means potential energy is maximum and kinetic energy is zero. 
Therefore, the correct statements are (a), (b), and (c) only.
Hence, option (2) is the correct answer.