Answer
$$\eqalign{
& \left( {\bf{a}} \right)\left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cr
& \left( {\bf{b}} \right)\left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr
& \left( {\bf{c}} \right) - {e^{ - 2t}} + 4t + {\sec ^3}t\tan t \cr
& \left( {\bf{d}} \right)4t{\sec ^2}t\tan t - 2{\sec ^2}t + 2{e^{ - t}}{\sec ^2}t\tan t + {e^{ - t}}{\sec ^2}t \cr
& {\text{ }} - 2{e^{ - t}} - 2t{e^{ - t}} \cr} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{e^{ - t}},{t^2},\tan t} \right\rangle \cr
& \left( {\bf{a}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {{e^{ - t}},{t^2},\tan t} \right\rangle } \right] \cr
& {\bf{r}}'\left( t \right) = \left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cr
& \cr
& \left( {\bf{b}} \right){\text{Find }}{\bf{r}}''\left( t \right) \cr
& {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle } \right] \cr
& {\bf{r}}''\left( t \right) = \left\langle {{e^{ - t}},2,2\sec t\sec t\tan t} \right\rangle \cr
& {\bf{r}}''\left( t \right) = \left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr
& \cr
& \left( {\bf{c}} \right){\text{Find }}{\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) \cr
& {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = \left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cdot \left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr
& {\bf{r}}'\left( t \right) \cdot {\bf{r}}''\left( t \right) = - {e^{ - 2t}} + 4t + {\sec ^3}t\tan t \cr} $$
\[\begin{gathered}
\left( {\mathbf{d}} \right){\text{Find }}{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - {e^{ - t}}}&{2t}&{{{\sec }^2}t} \\
{{e^{ - t}}}&2&{2{{\sec }^2}t\tan t}
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{2t}&{{{\sec }^2}t} \\
2&{2{{\sec }^2}t\tan t}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - {e^{ - t}}}&{{{\sec }^2}t} \\
{{e^{ - t}}}&{2{{\sec }^2}t\tan t}
\end{array}} \right|{\mathbf{j}} \hfill \\
{\text{ }} + \left| {\begin{array}{*{20}{c}}
{ - {e^{ - t}}}&{2t} \\
{{e^{ - t}}}&2
\end{array}} \right|{\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) = 4t{\sec ^2}t\tan t - 2{\sec ^2}t \cr
& {\text{ }} - \left( { - 2{e^{ - t}}{{\sec }^2}t\tan t - {e^{ - t}}{{\sec }^2}t} \right) + \left( { - 2{e^{ - t}} - 2t{e^{ - t}}} \right) \cr
& {\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) = 4t{\sec ^2}t\tan t - 2{\sec ^2}t + 2{e^{ - t}}{\sec ^2}t\tan t + {e^{ - t}}{\sec ^2}t \cr
& {\text{ }} - 2{e^{ - t}} - 2t{e^{ - t}} \cr
& \cr
& {\text{Summary}} \cr
& \left( {\bf{a}} \right)\left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cr
& \left( {\bf{b}} \right)\left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr
& \left( {\bf{c}} \right) - {e^{ - 2t}} + 4t + {\sec ^3}t\tan t \cr
& \left( {\bf{d}} \right)\left\langle { - \left( {t\cos t + \sin t} \right),\cos t - t\sin t,{t^2}} \right\rangle \cr} $$
$$\eqalign{
& {\text{Summary}} \cr
& \left( {\bf{a}} \right)\left\langle { - {e^{ - t}},2t,{{\sec }^2}t} \right\rangle \cr
& \left( {\bf{b}} \right)\left\langle {{e^{ - t}},2,2{{\sec }^2}t\tan t} \right\rangle \cr
& \left( {\bf{c}} \right) - {e^{ - 2t}} + 4t + {\sec ^3}t\tan t \cr
& \left( {\bf{d}} \right)4t{\sec ^2}t\tan t - 2{\sec ^2}t + 2{e^{ - t}}{\sec ^2}t\tan t + {e^{ - t}}{\sec ^2}t \cr
& {\text{ }} - 2{e^{ - t}} - 2t{e^{ - t}} \cr} $$