= Solution
Parallel <hyperbolic lines> have one common ideal endpoint and no interior intersection. Move the common endpoint to infinity by a <Möbius transformation>. The two lines are then distinct vertical lines $x=a,x=b$. Their <hyperbolic reflections> have product
$$
R_aR_b(z)=z+2(a-b),
$$
whose $m$th power is $z+2m(a-b)$. Since $a\ne b$, no positive power is the identity.
Ultraparallel <hyperbolic lines> have no common endpoint or intersection. To see their common perpendicular explicitly, first map one line to the imaginary axis. The other has two boundary endpoints on the same side, say $0<r<s$ after possibly reflecting. It is a semicircle of centre $(r+s)/2$ and radius $(s-r)/2$. The circle centred at zero with radius $\sqrt{rs}$ meets it orthogonally, since $((r+s)/2)^2=rs+((s-r)/2)^2$, and also meets the imaginary axis orthogonally. It supplies the common perpendicular. Map that perpendicular to the imaginary axis. Orthogonality then makes both lines semicircles centred at zero, of distinct radii $r,s$. This normal form can also be obtained by a real <Möbius transformation> carrying the endpoints of the common perpendicular to $0,\infty$. Their reflections are $z\mapsto r^2/\overline z$ and $z\mapsto s^2/\overline z$, so their product is
$$
z\mapsto\frac{r^2}{s^2}z.
$$
No positive power of this dilation is the identity because $r^2/s^2>0$ and $r\ne s$. Thus \b[the product has infinite order in both nonintersecting cases]. Geometrically the ultraparallel product is a translation along the common perpendicular through twice the separation of the lines.
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