I. solution. Between 6 AM and 6 PM, the hour hand is in every possible position exactly once, except for the straight-down position (6 o'clock). However, this is irrelevant from the perspective of the problem: at this time, it is visible that it is 6 o'clock, because if the hour hand were the vertically upward hand, the other would also be vertically upward.

For the clock with equally large hour and minute hands, the exact time cannot be read when it is equally possible that one hand is the minute hand and the other is the hour hand, but the two hands are not in the same position (in this case, it does not matter which is which).

Let's see when there is a situation where both positions are possible.

Let the rotation of one hand be $a$ degrees from 12 o'clock, in the clockwise direction. First, consider this as the hour hand. Then the other - the minute hand - has rotated $12a$ degrees, because in the same amount of time since the last midnight or noon, it has rotated 12 times as much. (Of course, $12a$ may include several 360-degree rotations.)

If we consider the other hand as the hour hand, we can similarly see that the first hand has rotated $144a$ degrees; its rotation can therefore be described by both $a$ and $144a$. Therefore:

$$
144a = a + k \cdot 360^\circ, \quad a = \frac{k \cdot 360^\circ}{143}
$$

This gives a different $a$ for each $k$ from 0 to 142, meaning that in the given time period, there are exactly 143 times when it was not possible to determine which hand was the hour hand and which was the minute hand.

Let's see how many times the two hands overlap! In this case, similarly:

$$
12a = a + k \cdot 360^\circ, \quad a = \frac{k \cdot 360^\circ}{11}
$$

So the two hands overlap in 11 cases.

There are 143 times when we cannot determine which hand is which; out of these, in 11 cases the two hands overlap, and in the remaining cases, they show different times if one hand is the hour hand and the other is the minute hand; therefore, there are exactly $\underline{132}$ cases where the exact time cannot be determined.

II. solution. Let's characterize the position of both hands by their rotation relative to the 12, using one twelfth of a full circle, $30^\circ$, as the unit. If in this system the hour hand is at position $a (0 \leq a < 12)$ - meaning that $a$ hours have passed since 0 or 12 o'clock - then the minute hand is at position $12\{a\}$, where $\{\ldots\}$ denotes the fractional part of the number. Thus, when the hour hand is at position $a$, the hands can be swapped to result in a possible position of the clock hands if and only if

$$
12\{12\{a\}\} = a
$$

To solve this equation, let's write the number $a$ in base 12. In this system, multiplying by 12 shifts the "twelfth-decimal" point to the right, and taking the fractional part removes the digits before the "twelfth-decimal" point.

Let $a = b_1, b_2 b_3 b_4 \ldots$, where $0 \leq b_i < 12$ are integers, the digits of the number. When calculating $12\{12\{a\}\}$ in base 12, we delete the first digit twice and shift the "twelfth-decimal" point to the right twice:

$$
12\{12\{a\}\} = b_3, b_4 b_5 \ldots
$$

If this number is equal to $a$, that is, $b_1, b_2 b_3 b_4 \ldots = b_3, b_4 b_5 \ldots$, then $b_1 = b_3, b_2 = b_4$, and the digits continue to repeat, so the number $a$ is of the form $a = b_1, b_2 \overline{b_1 b_2}$, and its base 12 representation is periodic with the two-digit $b_1 b_2$ group as the period.

Since $0 \leq b_1, b_2 < 12$, there are 144 different $b_1 b_2$ pairs and thus 144 such $a$ numbers. However, if $b_1 = b_2$, then the two hands overlap. In this case, we can naturally tell the time without having to distinguish the hands. This happens in 12 cases out of the 144 possibilities. (This actually represents only 11 points in time, since if $b_1 = b_2 = 11$, then the corresponding $11,111111 \ldots$ is the infinite twelfth fraction of 12, which corresponds to 0 o'clock.

In the remaining $144 - 12 = 132$ cases, however, the indistinguishable hands indeed make it impossible to determine whether

$$
a_{12} = b_1, b_2 b_1 b_2 \ldots \quad \text{or} \quad a_{21} = b_2, b_1 b_2 b_1 \ldots
$$

hours have passed.

Between 6 AM and 6 PM - and similarly in any 12-hour period - there are 132 times when the time cannot be determined if the hands cannot be distinguished.
How many times between 6 AM and 6 PM are there when we cannot figure out the exact time from the hour hand alone, if we cannot distinguish between the hour and minute hands?