We assume that the clock hands rotate at constant speed. That mathematical model does not describe all clocks well. In some clocks, the minute hand stays at say $17$ for almost $1$ minute, then moves very rapidly to $18$, with an irritating click.
1. For the first problem, it is clear that it will take a little more than an hour, say an hour plus $x$ minutes, where $x$ is well under $60$.
In $12$ hours the hour hand travels $360$ degrees. So it travels $30$ degrees per hour, and therefore $1/2$ degree per minute. In an hour and $x$ minutes the hour hand will have advanced by $30+x/2$ degrees.
In an hour the minute hand travels $360$ degrees, so it travels at $6$ degrees per minute. In an hour and $x$ minutes it will have travelled $360+6x$ degrees. So the minute hand will have advanced by $6x$ minutes. We therefore obtain the equation
$$30+\frac{x}{2}=6x.$$
Solve for $x$.
2. We sketch the interchangeability argument. Take some clock time $x$, where $x$ is measured in minutes from $12$:$00$. So for example $1$:$00$ o'clock is called $60$.
At time $x$, the hour hand is $x/2$ degrees clockwise from straight up. The minute hand is at $6x-360m$ degrees clockwise from straight up, for some integer $m$ chosen to make $6x-360m$ less than $360$ degrees.
Take another time $y$ minutes after straight up. Then the hour hand is at $y/2$ degrees from straight up, and the minute hand is at $6y-360n$ for some integer $n$.
Suppose that the hour and minute hands are identical in appearance. For us to be confused between $x$ and $y$, we must have $x\ne y$ and
$$\frac{x}{2}=6y-360 n \qquad \text{and} \qquad \frac{y}{2}=6x-360m.$$
We can use these equations to find all times $x\ne y$ such that times $x$ and $y$ are confused when the hands are identical, plus, of course, all times when the hands coincide.
But that's not what we want, since we were asked about $x-y$. Use the two equations to solve for this. We get $13(x-y)=720(m-n)$.
Note that we have measured the "times" $x$ and $y$ in minutes after straight up, since that's what we used in part $1$. Convert to minute spaces.
Hint: Try writing out formulae for the positions of the hands on a correct clock and one the wrong clock in terms of time.
Correct clock:
the long hand is at $(\text{time in mins})\times (6^{\circ}), \mod 360^{\circ}$, since it starts straight up and moves by 360 degrees each hour;
the short hand is at $180^{\circ} +(\text{time in mins})\times \frac{1}{2}^{\circ}, \mod 360^{\circ}$, since it starts straight down and moves by 360 degrees every 12 hours.
Now write similar equations for the wrong clock, and look for values of time from which the long-hand equations give the same angle and so do the short hand equations.
Best Answer
Every minute, the minute hand travels $6^{\circ}$. The hour hand travels $\frac1{12}$th as much, or $\frac12^{\circ}$.
If the minute hand is at $m$ and the hour hand is at $h$, we have $m+18 = h - \frac92 + 180$, or $m = h + 157.5^{\circ}$.
Additionally, we know it's past 3pm, so $h\geq 90^{\circ}$, and $h$ and $m$ have to be consistent - which means the minutes elapsed have to line up with where $h$ is. That is, $\frac{m}{6}\cdot\frac12+90=h$.
So we have two equations:
$$\begin{split} m &= h + 157.5 \\ \frac{m}{12} + 90 &= h \end{split}$$
Solving those gives us $m = 270^{\circ}, h = 112.5^{\circ}$. Which is 3:45pm.