1. Answer: 2 hours.
Let the distance between $B$ and $M$ be $S$ (km), then the distance between $A$ and $M$ is $2S$. Denote by $v$ (km/h) the speed of Ann and Bob, then the speed of Tom on a motorbike is $9v$.

Consider the case in which Tom first drives Ann. Since the distance between $A$ and $M$ is $2S$, and the rate of convergence of Ann and Tom is $v + 9v = 10v$, Ann and Tom will meet through $\frac{2S}{10v} = \frac{S}{5v}$ (h). After the same time Tom and Ann on a motorbike will drive from the meeting point to $M$. During this time equaled $2 \cdot \frac{S}{5v} = \frac{2S}{5v}$. Bob will pass the distance $v \cdot \frac{2S}{5v} = \frac{2S}{5}$. Therefore, when Tom with Ann will be in $M$, the distance between him and Bob will equal to $S - \frac{2S}{5} = \frac{3S}{5}$. Therefore, after leaving $M$ Tom will meet and pick up Bob in $\frac{3S}{5} : 10v = \frac{3S}{50v}$. The same time Tom (with Bob) will drive back to $M$. As a result, all three friends will be in $M$ in $\frac{2S}{5v} + 2 \cdot \frac{3S}{50v} = \frac{13S}{25v}$ (h) after the start of the motion.

Similarly, calculate the time in the case in which Tom first drives up Bob. Time of Tom on the way to a meeting with Bob and back to $M$ equals $\frac{S}{5v}$. During this time, Ann will pass the distance $\frac{S}{5}$. Therefore, when Tom (with Bob arrive) to $M$, Ann will be at a distance of $2S - \frac{S}{5} = \frac{9S}{5}$. Then Tom will need $2 \cdot \frac{9S}{5} : 10v = \frac{18S}{50v}$ for the road from $M$ to the meeting with Ann and back. As a result, in this case, all three friends will be in $M$ in $\frac{S}{5v} + \frac{18S}{50v} = \frac{14S}{25v}$ (h) after the start of the motion.

By condition, the total time for the whole path in the first case differs from the time in the second case by 2.4 minutes, i.e. by $\frac{1}{25}$ hours. Therefore:
$$\frac{14S}{25v} - \frac{13S}{25v} = \frac{1}{25},$$
whence $\frac{S}{v} = 1$. This means that Ann covers the distance $S$ by foot for 1 hour, so it takes her 2 hours for the distance from $A$ to $M$.