$4 \cdot 208$ At a certain time $T_{1}$, the hour hand is between 2 and 3, and the minute hand is between 4 and 5. After a certain period of time, at time $T_{2}$, the minute hand points to the position originally indicated by the hour hand, and the hour hand points to the position originally indicated by the minute hand. Try to find the original time $T_{1}$ and the current time $T_{2}$ in hours.
[Solution] Let $x$ be the degree of the hour hand at $T_{1}$, and $y$ be the degree of the minute hand at $T_{1}$ (as shown in the figure).

Since the hour hand moves $1^{\circ}$ in 2 minutes, and the minute hand moves $1^{\circ}$ in $\frac{1}{6}$ minutes, the required time is:
$T_{1}=2$ hours $\frac{y}{6}$ minutes, $T_{2}=4$ hours $\frac{x}{6}$ minutes.

We have the system of equations:
$\left\{\begin{array}{l}2(x-60)=\frac{1}{6} \cdot y, \\ 2(y-120)=\frac{1}{6} \cdot x .\end{array}\right.$
Solving these, we get $x=\frac{10080}{143}, y=\frac{18000}{143}$.
Thus, $\frac{x}{6}=\frac{1680}{143}=11 \frac{107}{143}, \frac{y}{6}=\frac{3000}{143}=20 \frac{140}{143}$.
Therefore, $T_{1}=2$ hours $20 \frac{140}{143}$ minutes, $T_{2}=4$ hours $11 \frac{107}{143}$ minutes (or approximately $T_{1}=2$ hours 20 minutes 59 seconds, $T_{2}=4$ hours 11 minutes 45 seconds).