(N/A) Given $\hat{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$ $(i)$ and $\hat{\theta} = -\sin \theta \hat{i} + \cos \theta \hat{j}$ (ii).
Multiplying $(i)$ by $\cos \theta$ and (ii) by $\sin \theta$ and subtracting: $\hat{r} \cos \theta - \hat{\theta} \sin \theta = (\cos^2 \theta + \sin^2 \theta) \hat{i} = \hat{i}$.
Multiplying $(i)$ by $\sin \theta$ and (ii) by $\cos \theta$ and adding: $\hat{r} \sin \theta + \hat{\theta} \cos \theta = (\sin^2 \theta + \cos^2 \theta) \hat{j} = \hat{j}$.
$(b)$ $|\hat{r}| = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1$ and $|\hat{\theta}| = \sqrt{(-\sin \theta)^2 + \cos^2 \theta} = 1$. Both are unit vectors.
$\hat{r} \cdot \hat{\theta} = (\cos \theta)(-\sin \theta) + (\sin \theta)(\cos \theta) = 0$. Thus,they are perpendicular.
$(c)$ $\frac{d\hat{r}}{dt} = \frac{d}{dt}(\cos \theta \hat{i} + \sin \theta \hat{j}) = -\sin \theta \frac{d\theta}{dt} \hat{i} + \cos \theta \frac{d\theta}{dt} \hat{j} = \omega(-\sin \theta \hat{i} + \cos \theta \hat{j}) = \omega \hat{\theta}$.
Similarly,$\frac{d\hat{\theta}}{dt} = \frac{d}{dt}(-\sin \theta \hat{i} + \cos \theta \hat{j}) = -\cos \theta \frac{d\theta}{dt} \hat{i} - \sin \theta \frac{d\theta}{dt} \hat{j} = -\omega(\cos \theta \hat{i} + \sin \theta \hat{j}) = -\omega \hat{r}$.
$(d)$ Given $\vec{r} = a\theta \hat{r}$. Since $\theta$ is dimensionless,$[r] = [a]$. Thus,$a$ has dimensions of length $[L]$.
$(e)$ $\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(a\theta \hat{r}) = a\dot{\theta}\hat{r} + a\theta\dot{\hat{r}} = a\omega\hat{r} + a\theta\omega\hat{\theta}$.
$\vec{a} = \frac{d\vec{v}}{dt} = a\dot{\omega}\hat{r} + a\omega\dot{\hat{r}} + a\dot{\theta}\omega\hat{\theta} + a\theta\dot{\omega}\hat{\theta} + a\theta\omega\dot{\hat{\theta}} = a\dot{\omega}\hat{r} + a\omega^2\hat{\theta} + a\omega^2\hat{\theta} + a\theta\dot{\omega}\hat{\theta} - a\theta\omega^2\hat{r} = (a\dot{\omega} - a\theta\omega^2)\hat{r} + (2a\omega^2 + a\theta\dot{\omega})\hat{\theta}$.