Answer
\[\begin{align}
& \left( \mathbf{a} \right)\left\langle t\cos t,t\sin t,1 \right\rangle \\
& \left( \mathbf{b} \right)\left\langle \cos t-t\sin t,\sin t+t\cos t,0 \right\rangle \\
& \left( \mathbf{c} \right)\text{ }t \\
& \left( \mathbf{d} \right)\left\langle -\sin t-t\cos t,\cos t-t\sin t,{{t}^{2}} \right\rangle \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \mathbf{r}\left( t \right)=\left\langle \cos t+t\sin t,\sin t-t\cos t,t \right\rangle \\
& \left( \mathbf{a} \right)\text{Find }\mathbf{r}'\left( t \right) \\
& \mathbf{r}'\left( t \right)=\frac{d}{dt}\left[ \left\langle \cos t+t\sin t,\sin t-t\cos t,t \right\rangle \right] \\
& \mathbf{r}'\left( t \right)=\left\langle -\sin t+t\cos t+\sin t,\cos t+t\sin t-\cos t,1 \right\rangle \\
& \mathbf{r}'\left( t \right)=\left\langle t\cos t,t\sin t,1 \right\rangle \\
& \\
& \left( \mathbf{b} \right)\text{Find }\mathbf{r}''\left( t \right) \\
& \mathbf{r}''\left( t \right)=\frac{d}{dt}\left[ \left\langle t\cos t,t\sin t,1 \right\rangle \right] \\
& \mathbf{r}''\left( t \right)=\left\langle \cos t-t\sin t,t\cos t+\sin t,0 \right\rangle \\
& \\
& \left( \mathbf{c} \right)\text{Find }\mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right) \\
& \mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right)=\left\langle t\cos t,t\sin t,1 \right\rangle \cdot \left\langle \cos t-t\sin t,t\cos t+\sin t,0 \right\rangle \\
& \mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right)=t{{\cos }^{2}}t-{{t}^{2}}\sin t\cos t+{{t}^{2}}\sin t\cos t+t{{\sin }^{2}}t+0 \\
& \mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right)=t{{\cos }^{2}}t+t{{\sin }^{2}}t \\
& \mathbf{r}'\left( t \right)\cdot \mathbf{r}''\left( t \right)=t \\
& \\
& \left( \mathbf{d} \right)\text{Find }\mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right) \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
t\cos t & t\sin t & 1 \\
\cos t-t\sin t & t\cos t+\sin t & 0 \\
\end{matrix} \right| \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix}
t\sin t & 1 \\
t\cos t+\sin t & 0 \\
\end{matrix} \right|\mathbf{i}-\left| \begin{matrix}
t\cos t & 1 \\
\cos t-t\sin t & 0 \\
\end{matrix} \right|\mathbf{j} \\
& \text{ }+\left| \begin{matrix}
t\cos t & t\sin t \\
\cos t-t\sin t & t\cos t+\sin t \\
\end{matrix} \right|\mathbf{k} \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=-\left( t\cos t+\sin t \right)\mathbf{i}+\left( \cos t-t\sin t \right)\mathbf{j} \\
& +\left( {{t}^{2}}{{\cos }^{2}}t+t\sin t\cos t-t\sin t\cos t+{{t}^{2}}{{\sin }^{2}}t \right) \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=-\left( t\cos t+\sin t \right)\mathbf{i}+\left( \cos t-t\sin t \right)\mathbf{j}+{{t}^{2}}\mathbf{k} \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left\langle -\left( t\cos t+\sin t \right),\cos t-t\sin t,{{t}^{2}} \right\rangle \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left\langle -\sin t-t\cos t,\cos t-t\sin t,{{t}^{2}} \right\rangle \\
\end{align}\]