Answer
$(a)\space \vec r=R(cos\theta\space i+sin\theta\space j)$
$(b)\space \theta=(\frac{2\pi}{T})\space t$
$(c)\space a=\frac{V^{2}}{R}$
Work Step by Step
Please see the attached image first.
(a) Let's assume point P(x,y) is the position of the particle at a certain time & we can write the position vector $(\vec r)$ as follows.
$\vec r=(OA)i+(AP)j= Rcos\theta\space i+Rsin\theta\space j$
$\vec r=R(cos\theta\space i+sin\theta\space j)$
(b) We know, that the rotation period (T) of the particle can express as follows.
$T=\frac{2\pi}{\omega}-(1)$
Also we know that the $\frac{\theta}{t}=\omega-(2)$
Let's apply the value of $\omega$ from equation (1) into (2). Then we get,
$\frac{\theta}{t}=\frac{2\pi}{T}=\gt\theta=(\frac{2\pi}{T})t$
(c)$\space \vec r=R(cos\theta\space i+sin\theta\space j)$
Let's differentiate this equation by t & we get the velocity of the particle.
$\frac{d}{dt}\vec r=R(\frac{d}{dt}cos\theta\space i+\frac{d}{dt}sin\theta\space j)$
$\vec V=R(-sin\theta\times\frac{d\theta}{dt}\space i+cos\theta\times\frac{d\theta}{dt}\space j)$
We know, $d\theta /dt=\omega,$ So we can get,
$\vec V=R\omega(-sin\theta\space i+cos\theta\space j)$
Now we can differentiate the above velocity vector by t & we get the acceleration vector of the particle.
$\frac{d}{dt}\vec V= R\omega(\frac{-d}{dt}sin\theta\space i +\frac{d}{dt}cos\theta \space j)$
$\vec a= \frac{d}{dt}\vec V= R\omega(-cos\theta\frac{d\theta}{dt}\space i -sin\theta\frac{d\theta}{dt} \space j)$
$\vec a= -R\omega^{2}(cos\theta\space i-sin\theta\space j)-(3)$
The magnitude of the acceleration$= \sqrt {(R\omega^{2}cos\theta)^{2}+(-R\omega^{2}sin\theta)^{2}}$
$=\sqrt {R^{2}\omega^{4}(cos^{2}\theta +sin^{2}\theta)}=R\omega^{2}$
We know $V=r\omega$ & we get,
$a=R(\frac{V}{R})^{2}=\frac{V^{2}}{R}$
