From the 7 male and 6 female sales representatives for an insurance company, a team of 2 men and 3 women will be selected to attend a national conference on insurance fraud.

In how many ways can the team of 5 be selected?
In how many ways can the team of 5 representatives be selected?

Step 1 ：The problem is asking for the number of ways to select a team of 5 representatives from a group of 7 men and 6 women, with the condition that the team must consist of 2 men and 3 women. This is a combination problem, as the order of selection does not matter.

Step 2 ：The number of ways to select 2 men from 7 is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of items, k is the number of items to select, and ! denotes factorial. Similarly, the number of ways to select 3 women from 6 is also given by the combination formula.

Step 3 ：Calculating the combinations, we find that there are \(C(7, 2) = 21\) ways to select the men and \(C(6, 3) = 20\) ways to select the women.

Step 4 ：The total number of ways to select the team is then the product of these two numbers, as each selection of men can be paired with each selection of women. So, the total number of ways is \(21 \times 20 = 420\).

Step 5 ：Final Answer: The team of 5 representatives can be selected in \(\boxed{420}\) ways.