Answer
(a)
The principal unit normal vector $N(t)$ is given by:
$$
\begin{aligned}
N(t)&=\langle 0, \cos t,-\sin t\rangle\\
\end{aligned}
$$
(b)
The curvature is given by:
$$
\begin{aligned}
k(t)=\frac{1}{5 t}.
\end{aligned}
$$
Work Step by Step
(a)
We have a parametrization
$$
r(t)= t^{2} \mathbf{i}+(\sin t-t \cos t) \mathbf{j}+( \cos t+t \sin t) \mathbf{k} \\
= \left\langle t^{2}, \sin t-t \cos t, \cos t+t \sin t\right\rangle
$$
$\Rightarrow $
$$
\begin{aligned}
r^{\prime}(t) &=\langle 2 t, \cos t+t \sin t-\cos t,-\sin t+t \cos t+\sin t\rangle\\
&=\langle 2 t, t \sin t, t \cos t\rangle
\end{aligned}
$$
$$
\begin{aligned}
\mathrm\Rightarrow\left|\mathrm{r}^{\prime}(t)\right| &=\sqrt{4 t^{2}+t^{2} \sin ^{2} t+t^{2} \cos ^{2} t} \\
& =\sqrt{4 t^{2}+t^{2}\left(\cos ^{2} t+\sin ^{2} t\right)} \\
&=\sqrt{5 t^{2}} \ \ \ \text {where} \ \ t \gt 0 \\
&=\sqrt{5} t
\end{aligned}
$$
Then, the unit tangent vector $T(t)$ is given by:
$$
\begin{aligned}
T(t) &=\frac{r^{\prime}(t)}{\left|r^{\prime}(t)\right|}\\
&=\frac{1}{\sqrt{5} t}\langle 2 t, t \sin t, t \cos t) \\
&=\frac{1}{\sqrt{5}}\langle 2, \sin t, \cos t\rangle
\end{aligned}
$$
Then
$$
\begin{aligned}
T^{\prime}(t)=\frac{1}{\sqrt{5}}\langle 0, \cos t,-\sin t\rangle
\end{aligned}
$$
$\Rightarrow$
$$
\left|\mathrm{T}^{\prime}(t)\right|=\frac{1}{\sqrt{5}} \sqrt{0+\cos ^{2} t+\sin ^{2} t}=\frac{1}{\sqrt{5}}
$$
Since the principal unit normal vector $N(t)$ is given by:
$$
N(t)=\frac{\mathrm{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|}
$$
Therefore,
$$
\begin{aligned}
N(t)&=\frac{\mathrm{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|} \\
&=\frac{1 / \sqrt{5}}{1 / \sqrt{5}}\langle 0, \cos t,-\sin t\rangle\\
& =\langle 0, \cos t,-\sin t\rangle\\
\end{aligned}
$$
(b)
The curvature is given by:
$$
\begin{aligned}
k(t)&=\frac{\left|\mathbf{T}^{\prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|} \ \ \text { by using the Formula 9} \\
& =\frac{1 / \sqrt{5}}{\sqrt{5} t}\\
&=\frac{1}{5 t}.
\end{aligned}
$$
