Answer
$${\bf{v}}\left( 4 \right) = {\bf{i}} + 8{\bf{j}} + 4{\bf{k}}$$$${\text{speed}} = 9$$$${\bf{a}}\left( 4 \right) = 2{\bf{j}} + {\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = t{\bf{i}} + {t^2}{\bf{j}} + \frac{1}{2}{t^2}{\bf{k}},{\text{ }}t = 4 \cr
& \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {t{\bf{i}} + {t^2}{\bf{j}} + \frac{1}{2}{t^2}{\bf{k}}} \right] \cr
& {\bf{v}}\left( t \right) = {\bf{i}} + 2t{\bf{j}} + t{\bf{k}} \cr
& {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {{\bf{i}} + 2t{\bf{j}} + t{\bf{k}}} \right\| \cr
& {\text{speed}} = \sqrt {1 + 4{t^2} + {t^2}} = \sqrt {1 + 5{t^2}} \cr
& {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {{\bf{i}} + 2t{\bf{j}} + t{\bf{k}}} \right] \cr
& {\bf{a}}\left( t \right) = 2{\bf{j}} + {\bf{k}} \cr
& \cr
& \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed at }}t = 4 \cr
& {\bf{v}}\left( 4 \right) = {\bf{i}} + 2\left( 4 \right){\bf{j}} + 4{\bf{k}} \cr
& {\bf{v}}\left( 4 \right) = {\bf{i}} + 8{\bf{j}} + 4{\bf{k}} \cr
& {\text{speed}} = \sqrt {1 + 5{{\left( 4 \right)}^2}} \cr
& {\text{speed}} = 9 \cr
& {\bf{a}}\left( 4 \right) = 2{\bf{j}} + {\bf{k}} \cr} $$
