= Solution
A unit-speed curve on a smooth embedded surface is a <geodesic> exactly when its acceleration is normal to the surface, equivalently when its tangential acceleration vanishes.
Every unit-speed geodesic on the cylinder through $(1,0,0)$ has the form
$$
\boxed{\gamma(s)=(\cos(as),\sin(as),bs)},
\qquad a^2+b^2=1.
$$
Indeed, unrolling the cylinder to its <universal cover> $\mathbb R^2$ turns these curves into straight lines. Directly,
$$
\gamma''(s)=-a^2(\cos(as),\sin(as),0),
$$
which is parallel to the cylinder's radial <normal vector>, verifying the geodesic characterization. Such a geodesic is closed exactly when $b=0$; its image is then the horizontal circle $z=0$.
\b[Yes.] In polar coordinates $(r,\theta)$ on $\mathbb R^2\setminus\{0\}$, use the <Riemannian metric>
$$
g=\frac{dr^2}{r^2}+d\theta^2.
$$
The coordinate $u=\log r$ identifies this surface isometrically with the flat cylinder $\mathbb R\times S^1$. Through each point, the circle $u=\text{constant}$ is a closed geodesic, and every other geodesic has nonzero linear motion in $u$ and is not closed. Thus every point lies on a unique closed geodesic.
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