How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels? Just a little bit unsure about this question:

How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?

And by 'words' I mean just a string of alphabet - does not have to be a real world in dictionary. I do know that you need to use Inclusion/Exclusion principle which says:

$|A \cup B| = |A| + |B| - |A\cap B|$

Let U be the set of all words containing 7 letters (26 letters to the power 7)
The only thing I'm not sure is what would be the sets A and B? Or do I need a third term C?
It would be easier to count the number of words that have $2$ or fewer vowels, and then subtract this number from the total number of words (which you have already computed).

I assume that by "$3$" or more vowels" you mean $3$ or more occurrences of vowels, so in particular a word with 2 e's, 2 i's, and the rest consonants qualifies.  

How many words are there with no vowels? Clearly 
$$21^7$$
 if, as per usual convention, we agree that there are $5$ vowels.

How many words with $1$ vowel?  Where the vowel occurs can be chosen in $\binom{7}{1}$ ways. For each of these ways, the vowel can be chosen in $5$ ways.  And once you have done that, the consonants can be filled in in $21^6$ ways, for a total of
$$\binom{7}{1}(5)(21^6)$$ 

Finally, how many with $2$ vowels? The location of the vowels can be chosen in $\binom{7}{2}$ ways.  Once this has been done, the actual vowels can be put into these places in $5^2$ ways.  And then you can fill in the consonants in $21^5$ ways, for a total of
$$\binom{7}{2}(5^2)(21^5)$$

Add up the $3$ numbers we have obtained, subtract from $26^7$.

Our argument was a little indirect. We could instead find, using the same sort of reasoning, the number of words with $3$ vowels, with $4$ vowels, with $5$, with $6$, with $7$, and add up.  This is only a little more work than the indirect approach.  But any saving of work is helpful!  Also, the indirect approach that was described lets us concentrate on pretty simple situations, the most complicated of which is the $2$ vowel case.

Remark: The calculation we have done is closely connected to the Binomial Distribution, and if you have already covered this, it may be the point of the exercise.  So if you know about the Binomial Distribution, imagine the letters to be chosen at random.  Then the number of patterns with $3$ or more vowels is the probability of $3$ or more vowels, multiplied by $26^7$.
Calculate: Let  = {, , ,…, , , , , } be the set of letters of the alphabet. Determine how many 5-letter anagrams can be formed with the letters of  so that the first and last letters are two distinct vowels and the remainder are three distinct consonants.