8. Figure 1 is a road map of a city, where $A$, $B, \cdots, I$ represent nine bus stops. A bus departs from station $A$, travels along the roads, reaches each bus stop exactly once, and finally returns to station $A$. The number of different sequences of stops it can pass through is


8. 32.

The bus route can be considered as a circle $\Gamma$ (direction is not considered for now), and each station on $\Gamma$ has exactly two adjacent stations.

Let the two stations adjacent to $I$ on $\Gamma$ be $\alpha$ and $\beta$. There are two cases:
(1) $\{\alpha, \beta\} \cap \{E, F, G, H\} \neq \varnothing$;
(2) $\{\alpha, \beta\} \cap \{E, F, G, H\} = \varnothing$ and they are not collinear.
For case (1), by symmetry, assume $\alpha = E$. If $\beta \neq A$ and $B$, then $A$ and $B$ are not adjacent to $I$. Thus, $A$ is adjacent to $E$ and $H$, and $B$ is adjacent to $E$ and $F$. This means $E$ has three neighbors, which is a contradiction.

Therefore, $\beta \in \{A, B\}$. Without loss of generality, assume $\beta = A$. In this case, $B$, $C$, and $D$ are not adjacent to $I$, so their neighbors on $\Gamma$ are already determined. Thus, we must have
$\Gamma = A I E B F C G D H A$.
Note that $\alpha$ has 4 choices, and $\beta$ has 2 choices, so there are $4 \times 2 = 8$ circles that satisfy case (1).

For case (2), assume $\alpha = A$ and $\beta = B$. In this case, $C$ and $D$ are not adjacent to $I$. Therefore, their neighbors on $\Gamma$ are already determined, and we have the segments $A I B$, $F C G D H$, and $E$ on $\Gamma$.
If $E$ is adjacent to $A$, then $B$ must be adjacent to $F$, giving us
$\Gamma = A I B F C G D H E A$.
Similarly, if $E$ is adjacent to $B$, then
$\Gamma = A I B E F C G D H A$.
If $E$ is not adjacent to $A$ or $B$, then $E$ must be adjacent to $F$ and $H$. In this case, the segments $A I B$ and $E F C G D H$ cannot be connected to form a circle.

Note that $\{\alpha, \beta\}$ has 4 choices, so there are $4 \times 2 = 8$ circles that satisfy case (2).

In summary, the number of circles corresponding to the bus route is 16, and each circle has two directions.
Therefore, the number of different orders in which the bus passes through the stations is $16 \times 2 = 32$.