The problem you are having with both set ups are that you are only considering one option. In the union question it says you have to have at least 4 union members. And you did the formula correctly for choosing 4 union members and 1 non-union. But you also have the option to have 5 union members. This is where the equation for 5 choose five comes from. When you have two different options for something then you need to add the possible combinations for each option.

IN the professor problem it says we have to have at least 1 professor which means you have to consider the situations with 1 professor, 2 professors and 3 professors and add each of them to get the total number of combination.

so you would have (4!/(4-1)!1!)(6!/(6-2)!2!) + (4!/(4-2)!2!)(6!/(6-1)!1!) + (4!/4-3!)3!) to get your answer.

In the quoted, 5 are selected from 13 ( 8 belong to the union, 5 do not) to form a group that has at least 4 belong to the union.
There're two cases:
4 from the union + 1 not from the union (in which no. of combination is 8C4*5C1)
5 all from the union ( in which no. of combination is 8C5)

In the "bold", 3 are selected from 10 ( 4 are professors, 6 are students) to form a group that has at least 1 professor.
There're such cases:
1 professor + 2 students : 4C1*6C2 = 4*15 = 60
2 professors + 1 student : 4C2*6C1 = 6*6 = 36
3 professors : 4C3 = 4
Total ways: 60+36+4 =100
However, there's another easier way to solve this. Find all possible combinations of 3 people, then eliminate combinations that consist no professor. We'll get combinations that have at least one professor.
pick 3 out of 6 students: 6C3 = 20
pick 3 out of total 10 people: 10C3 = 120
Total combinations that have at least one professor: 120 - 20 =100

professors - 4
students - 6

Committee must consist of 3 people.
If the committee has to include at least 1 professor
how many ways can this committee be formed?

Number of ways in which a committee can be formed =

3 Professors + 2 Professors, 1 Student + 1 professor, 2 Students

= 4C3 + 4C2*6C1 +4C1*6C2

=4 + 60 +36
= 100

The answer to this question will be 100.

However, there's another easier way to solve this. Find all possible combinations of 3 people, then eliminate combinations that consist no professor. We'll get combinations that have at least one professor.
pick 3 out of 6 students: 6C3 = 20
pick 3 out of total 10 people: 10C3 = 120
Total combinations that have at least one professor: 120 - 20 =100

how would you solve the " Union- employees " question on this method ?

There're ten persons, 4 professors and 6 students. We're asked to pick a group of three that has at least one professors.
then, however we pick, there will be only 4 cases that happen:
a. No professor, 3 students
b. 1 professor, 2 students
c. 2 professors, 1 student
d. 3 professors, no student
Note that the total combinations can be formed when we pick 3 out of 10,"10C3", is the sum of combinations for all a,b,c,d cases above.
Or a+b+c+d = 10C3
Now there're 2 approaches this case:
Direct approach : b+c+d = 4C1*6C2+4C2*6C1+4C3 = 60 + 36 + 4 = 100
Indirect approach: 10C3 - a = 10C3 - 6C3 = 120 - 20 = 100