Answer
$${\bf{v}}\left( 0 \right) = {\bf{i}} + {\bf{j}}$$$${\bf{a}}\left( 0 \right) = - \frac{1}{3}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = t{\bf{i}} + t{\bf{j}} + \sqrt {9 - {t^2}} {\bf{k}},{\text{ }}t = 0 \cr
& \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {t{\bf{i}} + t{\bf{j}} + \sqrt {9 - {t^2}} {\bf{k}}} \right] \cr
& {\bf{v}}\left( t \right) = {\bf{i}} + {\bf{j}} - \frac{t}{{\sqrt {9 - {t^2}} }}{\bf{k}} \cr
& {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {{\bf{i}} + {\bf{j}} - \frac{t}{{\sqrt {9 - {t^2}} }}{\bf{k}}} \right\| \cr
& {\text{speed}} = \sqrt {1 + 1 + \frac{{{t^2}}}{{9 - {t^2}}}} = \sqrt {\frac{{18 - 2{t^2} + {t^2}}}{{9 - {t^2}}}} \cr
& {\text{speed}} = \sqrt {\frac{{18 - {t^2}}}{{9 - {t^2}}}} \cr
& {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {{\bf{i}} + {\bf{j}} - \frac{t}{{\sqrt {9 - {t^2}} }}{\bf{k}}} \right] \cr
& {\bf{a}}\left( t \right) = - \frac{9}{{{{\left( {9 - {t^2}} \right)}^{3/2}}}}{\bf{k}} \cr
& \cr
& \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ at }}t = 0 \cr
& {\bf{v}}\left( 0 \right) = {\bf{i}} + {\bf{j}} - \frac{0}{{\sqrt {9 - {{\left( 0 \right)}^2}} }}{\bf{k}} \cr
& {\bf{v}}\left( 0 \right) = {\bf{i}} + {\bf{j}} \cr
& {\bf{a}}\left( 0 \right) = - \frac{9}{{{{\left( {9 - {{\left( 0 \right)}^2}} \right)}^{3/2}}}}{\bf{k}} \cr
& {\bf{a}}\left( 0 \right) = - \frac{1}{3}{\bf{k}} \cr} $$
