2.【Solution】The range of values for the numerator is from 1 to 5.

When the numerator is 1, the denominator can range from 2 to 59. There are 58 proper fractions. They are all irreducible fractions.

Since 2, 3, and 5 are all prime numbers, when the numerators are 2, 3, and 5, the denominators must and only need to satisfy the following two conditions:
1) The denominator is greater than the numerator and less than 60.
2) The denominator is not a multiple of the numerator.

After simple calculations, we can find:

When the numerator is 2, there are 29 suitable denominators;

When the numerator is 3, there are 38 suitable denominators;

When the numerator is 5, there are 44 suitable denominators;

Finally, consider the case when the numerator is 4, which is similar to when the numerator is 2. The denominator cannot be even, and it cannot be 3, so there are $28(=29-1)$ suitable denominators.

In summary, the total number of fractions that meet the requirements is: $58+29+38+44+28=197$ (fractions).

Answer: There are 197 irreducible proper fractions with a numerator less than 6 and a denominator less than 60.
2. How many irreducible proper fractions are there with a numerator less than 6 and a denominator less than 60?

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