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.
Your answer, (b), is correct, but the following may help clean up any problems with your reasoning:
You want to find out how many hours, $t$, elapse before the wall clock, $W_c$, and the table clock, $T_c$, show the same time again.
Since the wall clock gains $2$ minutes every $12$ hours, it really gains $\frac{t}{360}$ hours every $t$ hours (e.g., $\frac{12}{360}=\frac{1}{30}$ and $\frac{1}{30}$ hours is equivalent to $2$ minutes). Thus, we can let the total time elapsed on $W_c$ for $t$ hours be
$$
W_c = t+\frac{t}{360}.\tag{1}
$$
For the table clock, $T_c$, we actually lose $2$ minutes every $36$ hours; that is, we lose $\frac{t}{1080}$ hours every $t$ hours (for example, $\frac{36}{1080}=\frac{1}{30}$ and $\frac{1}{30}$ hours is equivalent to $2$ minutes, as described above). Thus, we can let the total time elapsed on $T_c$ for $t$ hours be
$$
T_c = t-\frac{t}{1080}.\tag{2}
$$
Now what? This depends on the kind of clock you are using. If you are using a regular wall-clock that does not differentiate between AM and PM (i.e., a 12-hour clock), then we will need to figure out when
$$
W_c-T_c=12.\tag{3}
$$
However, if you are using a clock that does differentiate between AM and PM, then you will need to figure out when
$$
W_c-T_c=24.\tag{4}
$$
Using a 12-hour clock: We substitute $(1)$ and $(2)$ into $(3)$ to get
$$
\frac{t}{360}+\frac{t}{1080}=12\Longleftrightarrow 4t=12960\Longleftrightarrow \color{red}{t=3240}.
$$
Thus, $3240$ hours will have elapsed. Note that
$$
\underbrace{3240}_{\text{hours}} = \underbrace{19}_{\text{weeks}}\cdot \underbrace{168}_{\text{hours/week}} + \underbrace{48}_{\text{hours}}.
$$
Thus, $19$ weeks and $48$ hours will have elapsed. Since your clocks began on a Tuesday at noon, they will next meet again on $\boxed{\color{red}{\text{Thursday at noon}}}$.
Using a 24-hour clock: We substitute $(1)$ and $(2)$ into $(4)$ to get
$$
\frac{t}{360}+\frac{t}{1080}=24\Longleftrightarrow 4t=25920\Longleftrightarrow \color{red}{t=6480}.
$$
Thus, $6480$ hours will have elapsed. Note that
$$
\underbrace{6480}_{\text{hours}} = \underbrace{38}_{\text{weeks}}\cdot \underbrace{168}_{\text{hours/week}} + \underbrace{96}_{\text{hours}}.
$$
Thus, $38$ weeks and $96$ hours (four days) will have elapsed. Since your clocks began on a Tuesday at noon, they will next meet again on $\boxed{\color{red}{\text{Saturday at noon}}}$.
Best Answer
If the minute hand has made $x$ revolutions around the clock ($60x$ minutes), the hour hand has made $\frac1{12}x$ revolutions; the difference between these two values represents an angle. Both hands start at 0 revolutions.
For (1), the relevant angle is $\frac14$ of the circle: $$x=\frac14+\frac1{12}x$$ Solving, we get $x=\frac3{11}$, which corresponds to a time of $\frac{180}{11}$ minutes after noon.
For (2) we replace $\frac14$ with 1, since the minute hand has lapped the hour hand. Then $$x=1+\frac1{12}x$$ Solving, we get $x=\frac{12}{11}$, i.e. $\frac{720}{11}$ minutes after noon.
Therefore, both your solutions are correct.