时钟问题：夹角与重合时间

(1) 3:15 时：

• 分针在 15 分处，对应角度 $15\times 6=90^\circ$；
• 时针在 3 点又走了 15 分钟，角度为 $3\times 30+15\times 0.5=90+7.5=97.5^\circ$。

夹角为 $|97.5-90|=\boxed{7.5^\circ}$。

(2) 设 3 点后经过 $m$ 分钟。

• 分针角度：$6m$；
• 时针角度：$90+0.5m$。

重合满足 $6m=90+0.5m$，即 $5.5m=90$，所以

$$m=\frac{180}{11}=16\frac{4}{11}\text{ 分钟}.$$

因此时间为 $\boxed{3:16:21\frac{9}{11}}$（约 3:16:21.818）。

(3) 同理，12 点后经过 $m$ 分钟，时针角度 $0.5m$、分针角度 $6m$，除 $m=0$ 外下一次重合满足 $6m=0.5m+360$，得到

$$5.5m=360\Rightarrow m=\frac{720}{11}=65\frac{5}{11}\text{ 分钟},$$

即 $\boxed{1:05:27\frac{3}{11}}$（约 1:05:27.273）。

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英文解析

(1) At 3:15:

• The minute hand is at the 15-minute mark, corresponding to an angle of $15\times 6=90^\circ$;
• The hour hand has moved past 3 o'clock by 15 minutes, with an angle of $3\times 30+15\times 0.5=90+7.5=97.5^\circ$.

The angle between them is $|97.5-90|=\boxed{7.5^\circ}$.

(2) Let $m$ minutes have passed after 3 o'clock.

• Minute hand angle: $6m$;
• Hour hand angle: $90+0.5m$.

Coincidence satisfies $6m=90+0.5m$, i.e., $5.5m=90$, so

$$m=\frac{180}{11}=16\frac{4}{11}\text{ minutes}.$$

Thus, the time is $\boxed{3:16:21\frac{9}{11}}$ (approximately 3:16:21.818).

(3) Similarly, let $m$ minutes have passed after 12 o'clock. The hour hand angle is $0.5m$ and the minute hand angle is $6m$. Excluding $m=0$, the next coincidence satisfies $6m=0.5m+360$, yielding

$$5.5m=360\Rightarrow m=\frac{720}{11}=65\frac{5}{11}\text{ minutes},$$

which is $\boxed{1:05:27\frac{3}{11}}$ (approximately 1:05:27.273).