Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{ Not continuous}} \cr
& \left( {\text{b}} \right){\text{ Continuous}} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right){\text{ }}{\bf{r}}\left( t \right) = {e^t}{\bf{i}} + {\bf{j}} + \csc t{\bf{k}} \cr
& {\text{Calculating if it is continuous at }}t = 0, \cr
& \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \mathop {\lim }\limits_{t \to 0} \left( {{e^t}{\bf{i}} + {\bf{j}} + \csc t{\bf{k}}} \right) \cr
& {\text{ }} = {\bf{i}}\mathop {\lim }\limits_{t \to 0} {e^t} + \mathop {\lim }\limits_{t \to 0} {\bf{j}} + {\bf{k}}\mathop {\lim }\limits_{t \to 0} \csc t \cr
& {\text{ }} = {\bf{i}}{e^0} + {\bf{j}} + {\bf{k}}\csc 0 \cr
& \mathop {\lim }\limits_{t \to 0} \csc t,{\text{ does not exist, then }}{\bf{r}}\left( t \right){\text{is not continuous at }}t = 0 \cr
& \left( {\text{b}} \right){\text{ }}{\bf{r}}\left( t \right) = 5{\bf{i}} - \sqrt {3t + 1} {\bf{j}} + {e^{2t}}{\bf{k}} \cr
& \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \mathop {\lim }\limits_{t \to 0} \left( {5{\bf{i}} - \sqrt {3t + 1} {\bf{j}} + {e^{2t}}{\bf{k}}} \right) \cr
& {\text{ }} = 5{\bf{i}}\mathop {\lim }\limits_{t \to 0} 1 - {\bf{j}}\mathop {\lim }\limits_{t \to 0} \sqrt {3t + 1} + {\bf{k}}\mathop {\lim }\limits_{t \to 0} {e^{2t}} \cr
& {\text{ }} = 5{\bf{i}} - {\bf{j}}\sqrt {3\left( 0 \right) + 1} + {\bf{k}}{e^{2\left( 0 \right)}} \cr
& {\text{ }} = 5{\bf{i}} - {\bf{j}} + {\bf{k}} \cr} $$
