Answer
$${\bf{v}}\left( 4 \right) = 8{\bf{i}} + {\bf{j}} + 6{\bf{k}}$$$${\bf{a}}\left( 4 \right) = 2{\bf{i}} - \frac{3}{4}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {t^2}{\bf{i}} + t{\bf{j}} + 2{t^{3/2}}{\bf{k}},{\text{ }}t = 4 \cr
& \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}{\bf{i}} + t{\bf{j}} + 2{t^{3/2}}{\bf{k}}} \right] \cr
& {\bf{v}}\left( t \right) = 2t{\bf{i}} + {\bf{j}} + 3{t^{1/2}}{\bf{k}} \cr
& {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {2t{\bf{i}} + {\bf{j}} + 3{t^{1/2}}{\bf{k}}} \right\| \cr
& {\text{speed}} = \sqrt {4{t^2} + 1 + 9t} \cr
& {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {2t{\bf{i}} + {\bf{j}} + 3{t^{1/2}}{\bf{k}}} \right] \cr
& {\bf{a}}\left( t \right) = 2{\bf{i}} - \frac{3}{2}{t^{ - 1/2}}{\bf{k}} \cr
& \cr
& \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ at }}t = 4 \cr
& {\bf{v}}\left( 4 \right) = 2\left( 4 \right){\bf{i}} + {\bf{j}} + 3{\left( 4 \right)^{1/2}}{\bf{k}} \cr
& {\bf{v}}\left( 4 \right) = 8{\bf{i}} + {\bf{j}} + 6{\bf{k}} \cr
& {\bf{a}}\left( 4 \right) = 2{\bf{i}} - \frac{3}{2}{\left( 4 \right)^{ - 1/2}}{\bf{k}} \cr
& {\bf{a}}\left( 4 \right) = 2{\bf{i}} - \frac{3}{4}{\bf{k}} \cr} $$
