Answer
$$r'(t) = \lt - \pi\sin\pi t, \pi\cos\pi t, 1 \gt, \quad T(t)= \frac{ \lt - \pi\sin\pi t, \pi\cos\pi t, 1 \gt}{\sqrt{ 1+\pi^2}}, \quad T(1)=\frac{ \lt 0, -\pi, 1 \gt}{\sqrt{1+\pi^2}}.$$
Work Step by Step
Recall that $(\sin x)'=\cos x$.
Recall that $(\cos x)'=-\sin x$.
Since $ r(t) = \lt \cos\pi t, \sin\pi t, t \gt$, then $ r'(t) = \lt - \pi\sin\pi t, \pi\cos\pi t, 1 \gt$ and $\|r'(t)\|=\sqrt{ 1+\pi^2} $.
Hence, we have
$$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt - \pi\sin\pi t, \pi\cos\pi t, 1 \gt}{\sqrt{ 1+\pi^2}}, \quad T(1)=\frac{ \lt 0, -\pi, 1 \gt}{\sqrt{1+\pi^2}}.$$
