13. Xiao Hua and Xiao Zhang are running at a constant speed on a circular track. They start at the same time and place. Xiao Hua runs clockwise, completing a lap every 72 seconds; Xiao Zhang runs counterclockwise, completing a lap every 80 seconds. A quarter-circle arc interval is marked on the track, centered at the starting point. How long, in seconds, do both of them stay in the designated interval at the same time?
【Analysis】(1) If Li's speed is $\frac{1}{72}$, the time to exit the designated area is $\frac{1}{8} \div \frac{1}{72}=9$ seconds. Zhang's speed is $\frac{1}{80}$, the time to exit the designated area is $\frac{1}{8} \div \frac{1}{80}=10$ seconds. Since 10 is greater than 9, the time is 9 seconds;
(2) The second time: Hua's time to enter the designated area is $\left(1-\frac{1}{4} \div 2\right) \div \frac{1}{72}=63$ (seconds), and the time to exit the area is $\left(1+\frac{1}{4} \div 2\right) \div \frac{1}{72}=81$ (seconds);
Zhang's time to enter the area is $\left(1-\frac{1}{4} \div 2\right) \div \frac{1}{80}=70$ (seconds), and the time to exit the area is $\left(1+\frac{1}{4} \div 2\right) \div \frac{1}{80}=90$ (seconds). The time they are in the designated area is from 70 to 81, which is 11 seconds;
(3) The third time: Hua's time is from 135 to 153, and Zhang's time is from 150 to 170. The overlapping time is from 150 to 153, which is 3 seconds;
(4) Hua's time to enter the designated area is $\left(1-\frac{1}{4} \div 2\right) \div \frac{1}{72}=63$ (seconds),

the time to exit the area is $\left(1+\frac{1}{4} \div 2\right) \div \frac{1}{72}=81$ (seconds);
$81-63=18$ (seconds);
Similar situations yield the same result.
【Solution】(1) Hua's time to exit the designated area is $\left(\frac{1}{4} \div 2\right) \div \frac{1}{72}=9$ seconds, and Zhang's time to exit the designated area is $\left(\frac{1}{4} \div 2\right) \div \frac{1}{80}=10$ seconds. Since $10>9$, the time is 9 seconds;
$$
\begin{array}{l}
\text { (2) } 80 \times\left(1-\frac{1}{4} \div 2\right), \\
=80 \times \frac{7}{8}, \\
=70 \text { (seconds); } \\
72 \times\left(1+\frac{1}{4} \div 2\right), \\
=72 \times \frac{9}{8}, \\
=81 \text { (seconds); } \\
81-70=11 \text { (seconds); }
\end{array}
$$
(3) The third time: Hua's time is from 135 to 153, and Zhang's time is from 150 to 170. The overlapping time is from 150 to 153, which is 3 seconds;
(4) Hua's time to enter the designated area is $\left(1-\frac{1}{4} \div 2\right) \div \frac{1}{72}=63$ (seconds),

the time to exit the area is $\left(1+\frac{1}{4} \div 2\right) \div \frac{1}{72}=81$ (seconds);
$81-63=18$ (seconds);
In summary: The answers are: $3,9,11,18$.
Answer: The time both are in the designated area is $3,9,11,18$ seconds.