Five, (20 points) A mall installs an escalator from the first floor to the second floor, moving at a uniform speed that is twice the speed of a child. It is known that a boy passed 27 steps to reach the top of the escalator, while a girl walked 18 steps to reach the top. (Assume the boy and girl each step on one step at a time)
(1) (10 points) How many steps are visible on the escalator?
(2) (10 points) If there is a staircase next to the escalator leading from the second floor to the first floor, with the same number of steps as the escalator, and the children each return to the top of the escalator at their original speeds, then walk down the stairs, and ride the escalator again (ignoring the distance between the escalator and the stairs), how many steps will the boy have walked when he catches up to the girl for the first time?
(1) Let the girl's speed be $x$ levels/min, the escalator's speed be $y$ levels/min, the boy's speed be $2x$ levels/min, and the stairs have $s$ levels, then
$$
\left.\begin{array}{l}
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y}
\end{array}\right. \\
\Rightarrow \frac{13.5}{18}=\frac{s-27}{s-18}
\end{array}\right\}
$$
(2) Let the boy pass the girl once after walking the escalator $m$ times and the stairs $n$ times, then the girl has walked the escalator $(m-1)$ times and the stairs at a speed of $x$ levels/min.
$$
\begin{array}{l}
\frac{54 m}{4 x}+\frac{54 n}{2 x}=\frac{54(m-1)}{3 x}+\frac{54(n-1)}{x}, \\
\frac{m}{4}+\frac{n}{2}=\frac{m-1}{3}+\frac{n-1}{1}, \\
6 n+m=16 .
\end{array}
$$
One of $m, n$ must be a positive integer, and $0 \leqslant|m-n| \leqslant 1$.
\begin{tabular}{c|ccccc}
$m$ & 1 & 2 & 3 & 4 & 5 \\
\hline$n=\frac{16-m}{6}$ & $2 \frac{1}{2}$ & $2 \frac{1}{3}$ & $2 \frac{1}{6}$ & 2 & $1 \frac{5}{6}$
\end{tabular}
\begin{tabular}{c|cc}
$n$ & 1 & 2 \\
\hline$m=16-6 n$ & 10 & 4
\end{tabular}

Obviously, only $m=3, n=2 \frac{1}{6}$,
So the boy walked the escalator 3 times and the stairs $2 \frac{1}{6}$ times.
Therefore, the total number of steps the boy walked $=3 \times 27+2 \frac{1}{6} \times 54=198$