= Solution
In the <Poincaré half-plane model>, $\mathbb H=\{x+iy:y>0\}$ has <Riemannian metric>
$$
\boxed{ds^2=\frac{dx^2+dy^2}{y^2}.}
$$
Its orientation-preserving <isometries> are the <Möbius transformations> $z\mapsto(az+b)/(cz+d)$ with real coefficients and $ad-bc=1$, with a simultaneous sign change representing the same map. The full <isometry group> also contains their compositions with $z\mapsto-\overline z$, which reverse <orientation>. The <hyperbolic lines> are vertical Euclidean lines and upper semicircles centred on the real axis; each meets the boundary orthogonally.
A <hyperbolic line> with finite endpoints $p<q$ is mapped to the <imaginary axis> by
$$
T(z)=\frac{z-p}{q-z}.
$$
Its coefficient <determinant> is $q-p>0$, so rescaling the coefficients makes this an orientation-preserving <isometry>. The endpoints map to $0$ and $\infty$, which characterize the imaginary-axis <hyperbolic line>. For a vertical line with endpoints $p,\infty$, use $T(z)=z-p$. If $T_1,T_2$ send the two prescribed lines to the <imaginary axis>, $T_2^{-1}T_1$ sends the first to the second. This proves <transitivity> on <hyperbolic lines>.
For two <ultraparallel hyperbolic lines>, choose their common perpendicular and map it to the <imaginary axis>. A <hyperbolic line> perpendicular to that axis is a semicircle centred at zero: <orthogonality> at $iy$ forces the <circle> centre to have real coordinate zero. Thus the reflecting lines become semicircles of distinct radii $r_1,r_2>0$. The <hyperbolic reflection> in the radius-$r$ semicircle is $S_r(z)=r^2/\overline z$; it fixes that semicircle pointwise and reverses <orientation>. Their composition is
$$
S_{r_2}S_{r_1}(z)=\lambda z,\qquad\lambda=(r_2/r_1)^2\ne1.
$$
Its $k$th power is $z\mapsto\lambda^kz$, and no positive power is the identity since $\lambda>0$ and $\lambda\ne1$. Conjugation by the chosen <isometry> preserves <order of a group element>, so \b[the original composition has infinite order]. Reversing the order of the reflections replaces $\lambda$ by $\lambda^{-1}$ and gives the same conclusion.
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