For $24$ hour time from $1:00$ to $11;59$ just mark $\text{am}$ to it to convert it into $12$ hour time.

To change a $pm$ time to $24$ hours time , you have to add $12 \ \ pm$ to the hours unless it is $12\ \ pm$ then the time remain unchanged .

To convert $24$ hour time from $13:00$ to $23:59$ into $12$ hour time just subtract $12$ from the hours and mark it as $\text{pm}$.

To change a $pm$ time to $24$ hours time , you have to add $12 \ \ pm$ to the hours unless it is $12\ \ pm$ then the time remain unchanged .

For $2$ hours time from $12:00$ to $12:59$ just maark $\text{pm}$ to it to convert it into $12$ hour time.

A day has $86400s=1440mins=24hrs$

The amount of degrees moved by the hour clock $=$
$=\dfrac{360}{12}(4+\dfrac{24}{60})$
$=120+12=132^\circ$

Assume that there is a line drawn between 12 and 6 on the clock, this can be a line of symmetry.
When we take the mirror image of this clock, the hands that are on the right side of this line will appear equidistant on the left side of the line and vice versa.
The time indicated is 6:10, the hour hand a little after 6 and the minute hand on 2.
Its mirror image will become, the hour hand a little before 6 and the minute hand on 10, i.e, 5:50.

The clock loses $5$ minutes per day
in 7 days, time lost $=5 \times 7=35mins$
actual time $=12'o clock$,
time shown $=11:25 a.m.$

The clock gains 12 mins in 12 hours,
It will gain 1 min in every 1 hour.
Difference in time the clock indicates=5:30 p.m.-11 a.m. (the next day)
=30.5 hours,
Time gained=30.5/1=30 mins.
Actual time=5 p.m.

In each hour, the time difference between the 2 clocks will increase by 5mins
time passed $=24+2+3=29hrs$ on clock 1
Real time on clock $1=4 p.m$
difference between 2 clocks=$5*30=150mins$,
time on clock $2=3hrs+2hrs+30mins$
$=5.30 p.m$

At twelve o'clock, the 2 hands are overlapping. 
As the minutes pass, the minutes hand moves farther away from the hours hand.
Therefore, between twelve o'clock and 1 o'clock, the hands will not overlap again.

Since the watch was 5 minutes fast on Wednesday evening, 5:00 p.m. and was 5 minutes slow after two days at 7:00 p.m., the correct time would be at the mid point of these 2 times.
Time difference=(24+24+2)50 hours,
So the right time will be 25 hours past 5 p.m. Wednesday,
I.e, 6 p.m. on Tuesday.

Let the hours passed be x,
time difference as per clock=5hrs and 30mins 
actual time=x*60+x*5=330,
x=5 hrs,
actual time=4:10

The clock will show 1 in an hour for 19 time for 11 hours it will show the incorrect time for $(19 \times 11)$ time. The last 12th hour will always show the in correct time so total in correct time.

$(19 \times 11 + 60)$ min = $269$ min

there are $24$ hours in a day to $ = 269 \times 2 = 538 $ min

$538$ min = $\displaystyle \frac {269} {30} = 9 $ hours

the fraction day when the clock shows the correct time is $\displaystyle = 1 - \frac {9} {24} $

                                                                                                 $\displaystyle = 1 - \frac {3} {8} = \frac {5} {8}$

After each hour, the difference between clocks will increase by 2 minutes.
At 10 p.m., hours passed=13,
Difference=13*2=26 minutes.

14:50 in the current clock would be indicated by the hour hand a little before midway of 7 and 8 and minute hand would be on 10.

Clock strikes the same number of times as the hour that it is in, i.e, once at 1. 
The hours on the clock range from 1 to 12.
In a span of 24 hours, it will complete the cycle twice.
number of times it strikes=2(1+2+...12)
=156

If you switch to a rotating coordinate system in which the hour hand stands still, then the minute hand makes only 11 revolutions, and so it is at right angles with the hour hand 22 times. In 36 hours, you get 322=66.

$The\quad clock\quad is\quad running\quad 6\quad minutes\quad slow\quad in\quad 1\quad day,\ %\quad of\quad time\quad lost=\frac { 6 }{ 24*60 } *100=\frac { 5 }{ 12 }%$

Required angle = 240 - 24 $\displaystyle \times $ (11/2)
                        = 240 - 132 = $\displaystyle 108^{\circ}$

Angle between hands of clock $=\left| 30H- \displaystyle\frac { 11 }{ 2 } M \right| $
where $H \rightarrow $ Hour hand, $M \rightarrow $ Minute hand
$\therefore  \left| 30\times 6- \displaystyle\frac { 11 }{ 2 } \times 30 \right| =15$

Let the hours passed be x,
time shown on clock 1=8:00 p.m. tuesday (24+24+10 hrs)
60x-2x=58*60,
x=60hrs,
actual time=10:00 p.m.
time on clock 2=(60-1)*60=59hrs,
=9:00 p.m. tuesday.

$12\rightarrow \frac { 360 }{ 12 } 30$

The two hands of a clock make $\displaystyle { 180 }^{ o }$ when they face each other in a straight line. 
This happens 11 times in 12 h. 
At 6 am and 6 pm, two hands form $\displaystyle { 180 }^{ o }$
Number of times they form $\displaystyle { 180 }^{ o }$ = 11 

Since the watch is  5 minutes slow at 12 noon on a Sunday and 5 minutes fast on the subsequent Wednesday at 6:00 p.m, it will show the right time at a time exactly mid way of theses 2 times.
hours passed=24+24+24+6=78hrs
mid way will be 39hrs after Sunday noon, which will be 3:00 a.m. tuesdsay.

When the hands are moving anticlockwise, we can assume that the time shown is the mirror image of actual time.
Assume that there is a line drawn between 12 and 6 on the clock, this can be a line of symmetry.
When we take the mirror image of this clock, the hands that are on the right side of this line will appear equidistant on the left side of the line and vice versa.
The time indicated is 8:25, the hour mid way after 8 and the minute hand on 5.
Its mirror image will become, the hour mid way before 4 and the minute hand on 7, i.e, 3:35.

In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.

To be together again, the minute hand must gain 60 minutes over the hour hand.

55 minutes are gained in 60 min.

60 min. are gained in [(60/55) * 60] min = $65\frac { 5 }{ 11 } $min.

Therefore, loss in 60 minutes =  $65\frac { 5 }{ 11 } -60=5\frac { 5 }{ 11 } $ min.

Loss in 24 hours = $5\frac { 5 }{ 11 } *\frac { 24*60 }{ 60 } $ = $130\frac { 10 }{ 11 } $min.

Therefore, the clock loses $130\frac { 10 }{ 11 } $ minutes in 24 hours.

Let the hours passed be $x$,
time shown on $clock 1=12:00 a.m$ saturday $(24*6+12hrs)$
$3600x-5x=156*60*60$,
$x=156.21hrs=156hrs$ and $13mins$
actual time $=12:13 a.m.$ saturday.

In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.

To be together again, the minute hand must gain 60 minutes over the hour hand.

55 minutes are gained in 60 min.

60 min. are gained in [(60/55) * 60] min = $65\dfrac { 5 }{ 11 } $min.

But they are together after 65 min.

Therefore, gain in 65 minutes =  $65\dfrac { 5 }{ 11 } -60=\dfrac { 5 }{ 11 } $ min.

Gain in 24 hours = $\dfrac { 5 }{ 11 } * \dfrac { 24*60 }{ 65 } $ = 1440/143 min.

Therefore, the clock gains $(10 + 10/143 )$ minutes in $24$ hours.

$2$ hours before midnight means $22:00$.