Amy and Beatrice watches

For each hour in Beatrice's watch,it was 61 minutes in Amy's watch;for every hour in actual time,it was * $\frac{61*58}{60}$ * that is 58 $\frac{58}{60}$ minutes in Amy's watch.This means that Amy's watch is slower than actual time by 1 minute and 2 seconds every hour.For two hours in actual time, Amy's watch was slower by 2 minutes and 4 seconds .Thus the watch showed *10:57:56 so abcd=5756( is times):) **

After an hour of actual time has lapsed, Beatrice's watch would show 58 minutes. Her watch movement rate is $\omega_B = \frac {58}{60}$ on actual time. When Beatrice's watch shows 1 hour or 60 minutes, Amy's watch shows 61 minutes. The movement rate of Amy's watch is then $\omega_{AB} = \frac {61}{60}$ on Beatrice's time. Therefore, the movement rate of Amy's watch on actual time is given by $\omega_A = \omega_AB \omega_B= \frac {61\times 58}{60\times 60}$ .

After two hours of actual time has lapsed, the time shown on Amy's watch is:

$\begin{aligned} t & = \frac {61\times 58}{60 \times 60} \times 2 \times 60 \times 60 \text{ s} \\ & = \text{0:0:7076} = \text{0:117:56} = \text{1:57:56} \end{aligned}$

Therefore the time shown on Amy's watch is $\text{10:57:56}$ , then $\overline{abcd} = \boxed{5756}$ .

$2\text{ hrs} \times \frac{60 \text{ min}}{1\text{ hr}} \times \frac{58 \text{ Beatrice min}}{60 \text{ min}} \times \frac{61 \text{ Amy min}}{60 \text{ Beatrice min}} = \frac{1769}{15}\text{ Amy min} = 1\!:\!57\!:\!56 \text{ Amy}$

Thus, Amy's watch will show $9\!:\!00\!:\!00 + 1\!:\!57\!:\!56 = 10\!:\!\boxed{57\!:\!56}$ .