There's a curve $r(t)=(\cos t, \sin t, t)$ and a body that is traveling the curve. At $t=\pi$ the body leaves the curve. Given that no external forces influence the body, where (which point) will it be situated at $t=2\pi$?
So we can suppose that the body will just travel in the direction of the tangent once it leaves the curve. We can find this direction: $$ r'(t)=<-\sin t, \cos t, 1> $$ $$ r'(t=\pi)=<0,-1,1> $$
My next step would be to find the line with the same direction vector as $r'(t)$, so we need to find the starting point which is $r(t=\pi)=(-1,0,\pi)$ so now we have the line: $$ l: (-1,0,\pi)+t(0,-1,1) $$
So now we can just plug in $2\pi$ into line equation to get the point where the body will: $$ 2\pi(0,-1,1)=(0,-2\pi,2\pi) $$
I don't think the solution is correct and I feel that unit vector tangent is involved somewhere.
You are almost there. Only the $t$-parametrization of $l$ is slightly off the mark. It should be $$l:\qquad t\mapsto l(t):=r(\pi)+(t-\pi)r'(\pi)$$ instead. Note that at time $t=\pi$ the running point on $l$ starts at $r(\pi)$ and proceeds with constant speed $r'(\pi)$ along $l$. It follows that $$l(2\pi)=(-1,0,\pi)+\pi(0,-1,1)=\ldots$$