Uniform Circular Motion

In uniform circular motion, the speed is constant but the direction of velocity changes continuously; since velocity is a vector, it is varying. The centripetal acceleration has constant magnitude $\frac{v^2}{r}$ but its direction (towards the centre) keeps changing, so acceleration is also varying. So the particle has varying velocity and varying acceleration.

$x^2 + y^2 = 16$, so the path is a circle of radius 4 m, and the acceleration is centripetal (along $-\vec R$) with magnitude $\frac{v^2}{R}$.
$\vec v = \frac{d\vec R}{dt} = 8\pi\cos 2\pi t\,\hat i - 8\pi\sin 2\pi t\,\hat j$
$|\vec v| = 8\pi$ m/s, not 8 m/s — so that statement is wrong.

R = 5 cm = $5\times10^{-2}$ m, T = $0.2\pi$ s.
Angular speed $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2\pi} = 10$ rad/s
For constant speed the only acceleration is centripetal: $a = \omega^2R$
$a = (10)^2\times5\times10^{-2} = 100\times0.05$
a = 5 m/$s^2$

Frequency: $n = \dfrac{22}{44} = \dfrac{1}{2}$ revolution per second
Angular speed: $\omega = 2\pi n = 2\pi \times \dfrac{1}{2} = \pi$ rad/s
Centripetal acceleration: $a_r = \omega^2 r = (2\pi n)^2 r$
$a_r = (\pi)^2 \times 1 = \pi^2$ m s$^{-2}$
The speed is constant, so the tangential acceleration is zero: $a_T = 0$.
So the acceleration is $\pi^2$ m s$^{-2}$, directed along the radius towards the centre.

Angular velocity: $\omega = \dfrac{2\pi}{T}$
It depends only on the time period, not on the mass or the radius.
Both particles have the same time period T, so $\omega_1 = \omega_2$.
$\dfrac{\omega_1}{\omega_2} = 1$
(Their linear speeds $v = \omega r$ are different, in the ratio R : r.)