8. A scientist invented a time machine, which looks like a circular subway track. Now (2010) is the first platform, the 2nd, 3rd, ..., 2009th platforms are 2011, 2012, ..., 4018, and the 2010th platform returns to the present (the departure platform). Later, the machine had a programming error, changing its operation rules to: the passenger specifies a time (i.e., platform number), the machine first arrives at the specified platform, then stops every 4 stations, at the 5th station. If the platform number where it stops is a positive integer power of 2, it moves back 2 stations to stop (e.g., $17 \rightarrow 22 \rightarrow 27 \rightarrow 32 \rightarrow 30 \rightarrow 35 \rightarrow \cdots$); if it stops at the first platform, it stops working. Try to answer:
(1) Can this machine get lost in the time track and fail to return to the present (i.e., not stop at the first platform)?
(2) If it can eventually return to the present, what is the maximum number of platforms the machine can stop at?
8. (1) The time track has 2009 stations, and the positive integer powers of 2 not exceeding 2009 are
$$
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024.
$$

According to the rules,
$$
\begin{array}{l}
\text { (i) (1) } 5 k+1(k=1,2,3) \\
\xrightarrow{3-k \text { times }} 16 \rightarrow 14=5 \times 2+4 \text {; } \\
\end{array}
$$
(2) $5 k+1(k=4,5, \cdots, 51)$
$$
\xrightarrow{51-k \text { times }} 256 \rightarrow 254=5 \times 50+4 \text {; }
$$
(3) $5 k+1(k=52,53, \cdots, 401)$
$$
\xrightarrow{401-k \text { times }} 2006 \rightarrow 2 \text {. }
$$
(ii)(1) $2 \rightarrow 2009 \rightarrow 5$;
(2) $5 k+2(k=1,2, \cdots, 6)$
$$
\xrightarrow{6-k \text { times }} 32 \rightarrow 30=5 \times 6 \text {; }
$$
(3) $5 k+2(k=7,8, \cdots, 102)$
$$
\xrightarrow{102-k \text { times }} 512 \rightarrow 510=5 \times 102 \text {; }
$$
(4.) $5 k+2(k=103,104, \cdots, 401)$
$$
\xrightarrow{401-k \text { times }} 2007 \rightarrow 3 \text {. }
$$
(iii)(1) $3 \rightarrow 8 \rightarrow 6=5+1$;
(2) $5 k+3(k=1,2, \cdots, 25)$
$$
\xrightarrow{25-k \text { times }} 128 \rightarrow 126=5 \times 25+1 \text {; }
$$
(3) $5 k+3(k=26,27, \cdots, 401)$
$$
\xrightarrow{401-k \text { times }} 2008 \rightarrow 4 \text {. }
$$
(iv) (1) $4 \rightarrow 2 \rightarrow 2009 \rightarrow 5$;
(2) $5 k+4(k=1,2, \cdots, 12)$
$$
\xrightarrow{12-k \text { times }} 64 \rightarrow 62=5 \times 12+2 \text {; }
$$
(3) $5 k+4(k=13,14, \cdots, 204)$
$$
\xrightarrow{204-k \text { times }} 1024 \rightarrow 1022=5 \times 204+2 \text {; }
$$
(4. $5 k+4(k=205,206, \cdots, 401)$
$$
\xrightarrow{401-k \text { times }} 2009 \rightarrow 5 \text {. }
$$
(v) $5 k(k=1,2, \cdots, 401)$
$$
\xrightarrow{401-k \text { times }} 2005 \rightarrow 1 \text {. }
$$

Let the passenger's designated station number be $a(1<a \leqslant 2009)$.
From (i) to (v), we know that no matter what $a$ is, the machine can always return to the present.
(2) Let the track from station 2009 to station 1 be denoted as $A$.

Let the number of times the machine passes through $A$ be $s$, and the total number of stations it stops at be $t$, where it stops at stations with numbers that are powers of 2 $v$ times, and at other stations $u$ times. Then
$$
\begin{array}{l}
2009 s+1=a+5(u-1)-2 v \\
=a+5 t-5-7 v . \\
\text { Hence } t=\frac{2009 s+7 v-a+6}{5} .
\end{array}
$$

We now prove that the machine will not stop at the same station twice.

Otherwise, the sequence of stations the machine stops at would form a loop. The result would be that it would never stop at the first station (contradicting (1)), or it would have already stopped at the first station in the first loop segment, thus stopping and not forming a loop, which is a contradiction.
Therefore, $v \leqslant 10$.
Consider the station number the machine stops at first after each pass through $A$, which is contained in $\{1,2,3,4,5\}$.
Notice that,
$3 \rightarrow 8 \rightarrow 6 \xrightarrow{2 \text { times }} 16 \rightarrow 14 \xrightarrow{10 \text { times }} 64 \rightarrow 62$
$\xrightarrow{90 \text { times }} 512 \rightarrow 510 \xrightarrow{299 \text { times }} 2005 \rightarrow 1$,
$4 \rightarrow 2 \rightarrow 2009 \rightarrow 5 \xrightarrow{400 \text { times }} 2005 \rightarrow 1$.
Thus, the machine has stopped at station 3, or at stations 4, 2, 5, but not both.
Therefore, $s \leqslant 2$.
If $s=1$, then $t \leqslant \frac{2009+70-0+6}{5}=417$.
If $s=2$, then the second loop is
$3 \xrightarrow{407} 1$ times or $2 \xrightarrow{403 \text { times }} 1$, or $4 \xrightarrow{404} 1$.
We now use (1) (i) to (v) to trace back the first loop from 3 or 2, 4:
$$
\begin{array}{l}
3 \leftarrow 2007 \stackrel{197 \text { times }}{\longleftarrow} 1022 \leftarrow 1024 \stackrel{154 \text { times }}{\longleftarrow} \\
254 \leftarrow 256 \longleftarrow 26 \text { times } 126 \leftarrow 128 \longleftarrow 23 \text { times } 13 \text {, } \\
2 \leftarrow 2006 \stackrel{349 \text { times }}{\longleftarrow} 261 \text {, } \\
4 \leftarrow 2008 \stackrel{375 \text { times }}{\longleftarrow} 133 \text {. } \\
\end{array}
$$

In summary, if $a=13$, then $v=7, t=812$;
if $a=261$, then $v=1, t=754$;
if $a=133$, then $v=2, a=781$.
Therefore, the machine can stop at a maximum of 812 stations.