Answer
$ a.\quad$ see image
$ b.\quad \mathrm{r}^{\prime}(t) =(\cos t) \mathrm{i}-(2\sin t)\mathrm{j}$,
$ c.\quad$ see image
Work Step by Step
$a.$
Parametric equations: $\mathrm{r}(t):\quad\left\{\begin{array}{ll}
x=\sin t & \\
y=2\cos t &
\end{array}\right.$
Apply $ \sin^{2}A+\cos^{2}A=1:$
halve y and square it $\displaystyle \Rightarrow (\frac{y}{2})^{2}=\cos^{2}t...$
we obtain:
$x^{2}+\displaystyle \frac{y^{2}}{4}=1,\qquad $... an ellipse.
see image
$b.$
Differentiate $(\displaystyle \frac{d}{dt})$ each component function
$\mathrm{r}^{\prime}(t):\quad\left\{\begin{array}{l}
x=\cos t\\
y=-2\sin t
\end{array}\right.$
$\mathrm{r}^{\prime}(t) =\langle\cos t,-2\sin t\rangle$
$\mathrm{r}^{\prime}(t)=(\cos t) \mathrm{i}-(2\sin t)\mathrm{j}$,
$c. $
For t = $\pi/4,$
position vector : black$\quad \mathrm{r}(\pi/4)$=$\displaystyle \langle\frac{\sqrt{2}}{2},\sqrt{2}\rangle$
tangent vector: red$\quad \mathrm{r}^{\prime}(\pi/4)$=$\displaystyle \langle\frac{\sqrt{2}}{2},-\sqrt{2}\rangle$
