(1)

Given, $\overrightarrow{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}$

$\therefore \overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=-\omega \sin \omega t \hat{x}+\omega \cos \omega t \hat{y}$

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=-{\omega }^2\cos \omega t \hat{x}-{\omega }^2\sin \omega t \hat{y}=-{\omega }^2\overrightarrow{r}$

Since position vector $\left(\overrightarrow{r}\right)$ is directed away from the origin, so, acceleration $(-{\omega }^2\overrightarrow{r})$ is directed towards the origin.

Also, 

$\overrightarrow{r}·\overrightarrow{v}=(\cos \omega t \hat{x}+\sin \omega t \hat{y})·(-\omega \sin \omega t \hat{x}+\omega \cos \omega t \hat{y})$

$=-\omega \sin \omega t\cos \omega t+\omega \sin \omega t\cos \omega t=0$

$\Rightarrow \overrightarrow{r}\perp \overrightarrow{v}$

(1)

Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are orthogonal to each other, if their scalar product is zero i.e. $\overrightarrow{A}·\overrightarrow{B}=0$

Here, $\overrightarrow{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$

and $\overrightarrow{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}$

$\therefore$   $\overrightarrow{A}·\overrightarrow{B}=(\cos \omega t \hat{i}+\sin \omega t \hat{j})·(\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j})$

        $=\cos \omega t\cos \frac{\omega t}{2}+\sin \omega t\sin \frac{\omega t}{2}=\cos (\omega t-\frac{{\displaystyle \omega t}}{{\displaystyle 2}})$

But $\overrightarrow{A}·\overrightarrow{B}=0$ (as $\overrightarrow{A}$ and $\overrightarrow{B}$ are orthogonal to each other)

$\therefore$   $\cos (\omega t-\frac{\omega t}{2})=0$

        $\cos (\omega t-\frac{\omega t}{2})=\cos \frac{\pi }{2} \mathrm{or} \omega t-\frac{\omega t}{2}=\frac{\pi }{2}$

        $\frac{\omega t}{2}=\frac{\pi }{2} \mathrm{or} t=\frac{\pi }{\omega }$