Hour, second and minute hands

Define $e(t):= t \>{\rm mod}\>1$. The orbit ${\bf x}$ of the three hands is then given by $${\bf x}:\quad t\mapsto\bigl(e(t),e(12t),e(720t)\bigr)\ .$$ Instead we look at the orbit ${\bf x}'=(x_1,x_2)$ of the second and third hands relative to the first hand, given by $${\bf x}': \quad t\mapsto\bigl(e(11t),e(719t)\bigr)\ .$$ In the $(x_1,x_2)$-plane we draw the fundamental domain $R:=\bigl[-{1\over2},{1\over2}\bigr]\times\bigl[-{1\over2},{1\over2}\bigr]$. It is easily seen that the part $F:$ $|x_1-x_2|\leq{1\over2}$ of $R$ (a hexagon) contains the positions where the three hands can be covered by a half-disk. This part $F$ makes up ${3\over4}$of $R$.

If we now draw the relative orbit ${\bf x}'$ into $R$ then we see a "lattice" of $719$ parallel lines with slope ${719\over11}$ in $R$. The wandering point ${\bf x}'(t)$ has constant speed. It is then clear that ${\bf x}'(t)$ spends $\approx{3\over4}$ of its time in $F$. In order to obtain the exact (rational) value one would have to engage in cumbersome calculations, which I omit.

(Many thanks to Empy2 and gandalf61 for spotting a mistake in the first version! Hopefully they won't find another one :)

Not an answer, just a ballpark estimate:

In[141]:= hourAngle[h_, m_, s_] := 2*Pi*(h*3600 + m*60 + s)/43200;
minuteAngle[h_, m_, s_] := 2*Pi*(m*60 + s)/3600;
secondAngle[h_, m_, s_] := 2*Pi*(s)/60;
sameSemiCircle[h_, m_, s_] := Module[
   {a1, a2, a3, a1s, a2s, a3s, a1norm, a2norm, a3norm},
   a1 = hourAngle[h, m, s];
   a2 = minuteAngle[h, m, s];
   a3 = secondAngle[h, m, s];
   {a1s, a2s, a3s} = Sort[{a1, a2, a3}];
   a1norm = 0;
   a2norm = a2s - a1s;
   a3norm = a3s - a1s;
   If[a2norm >= Pi || Not[Pi < a3norm < (a2norm + Pi)], True, False]
   ];

In[145]:= count = 0;
For[i = 0, i <= 11, i++,
  For[j = 0, j <= 59, j++,
    For[k = 0, k <= 59, k++,
      If[sameSemiCircle[i, j, k], count++];
      ];
    ];
  ];
count

Out[147]= 32406

Basically, I have checked all times from 00:00:00 to 11:59:59 with one second step and checked if clock hands are in the same semi-circle. I know this is an approximation because my time is descrete, but it should give quite good estimate.

It turns out, hands are in the same semi-circle for 28516 seconds out of 43200 or approx. 75.01% of time. This looks suspiciously close to 3/4 and it could very well be the final answer.

A word about the script: for any given time it's easy to calculate angles between noon and clock hands. I have sorted that angles and normalized them by subtracting the smallest one from the remaining two. Normalized angles are, say, $a_1=0\le a2 \le a3 \lt 2\pi$. It's like having one hand fixed at noon position.

You have two possible situations:

1. If $a_2<\pi$, the third angle must be outside of the shaded zone if we want all three hands to be in the same semicircle
2. If $a_2\ge\pi$, all three hands are guaranteed to be in the same semicircle, because $a_3\ge a_2$

That is where the key line of code comes from:

If[a2norm >= Pi || Not[Pi < a3norm < (a2norm + Pi)], True, False]

Monte Carlo method with 2 billion randomly chosen (not descrete) times gives 75.003% up to three decimal places.