x amount of people drink coffee, y amount of people drink tea, z amount of people drink both 
There are coffee drinkers and there are tea drinkers and among them
  there are those who like both.  Among the management students it was
  found that the probability that the student drinks coffee is 0.32, the
  probability that the student drinks tea is 0.45,  and the probability
  that the student drinks both is 0.26.
What is the probability that the student doesn't drink tea?

if .45 of people drink tea then 1-.45 of people don't drink tea?

What is the probability that the student drinks neither coffee nor tea?

.45+.32 people drink tea or coffee so 1 - .45+.32 drink neither?

What is the probability that the student drinks coffee or tea?

.45+.32 people drink coffee or tea?
 A: It may help to draw a Venn diagram and fill in the parts you know:

You know the entire content of the "Coffee" circle, the "Tea" circle and also the overlap. You know the probability of the entire box is 1. 
From those you can work out the content of each "piece" (Coffee-only, Tea-only, Neither and you already have Both), like so:

and from those pieces all the questions are immediately answered.
A: What is the probability that the student drinks neither coffee nor tea?
I will calculate the % of people who just drink tea, and just drink coffee.
  drinks tea is 0.45
-drinks both is 0.26
====================
drinks just tea 0.19 


  drinks coffee is 0.32
   -drinks both is 0.26
=======================
drinks just coffee 0.06

Then the problem is much less tricky:
What is the probability that the student drinks neither coffee nor tea?
1 (All) - .06 (Cof) -.19 (Tea) - .26 (Both) = .49
Feel free to ask questions.  Sorry about any formatting problems.
Once you get the hang it this, you can jump to:
1-.45-.32+.26=.49 
or 1 - (.45 Tea/Both +.32 Coff/Both -.26 Both) The minus both is because both would have gotten counted twice.
That is really the main trick of the problem, you must not count the both group twice.
