The word "either" typically denotes exclusivity of a logical disjunction. In other words, given statements $x_1,...,x_n$, the following are equivalent:
(1) Either $x_1$ or ... or $x_n$ holds.
(2) Exactly one of the statements $x_1,...,x_n$ holds.
(3) (At least / No less than) one of the statements $x_1,...,x_n$ holds, and (at most / no more than) one of them holds.
Unfortunately, the word "either" is often misused, even by people (e.g.: English-speaking mathematicians) who really ought to know better! Some use it simply as a lead-off word for a list of options, and in that case, it is effectively a throwaway word, meaning nothing at all, and so allowing that any non-null subcollection of the list of statements may hold simultaneously.
As a further complication, the phrase "any one of" seems to carry with it the connotation of "exactly one of"! It would be better to say "any of" instead--as that suggests that more than one may hold, but at least one must hold--but again, this is a common error, even among those who should know better.
Ultimately, this is just a frustrating reality. The best advice I can offer is to try to determine which meaning is intended by the questioner. If it is a textbook, try to find an early example where "either" is used, and see if it means "exactly" or if it is a throwaway. Likewise with "any one of". It is probable that they are intended to have different meanings, so if one is exclusive, then the other is almost surely inclusive.
Edit: Based on what you've told me, it looks as though the author is using "either" as a throwaway word, which suggests that "any one of" is intended to mean "exactly one of" in this text.
Your approach to the first problem is then the right one to take. I am going to break things down as follows, to give a slightly more general approach that should help out in both problems and more, if you're familiar with the basic concepts of set theory (specifically: union, intersection, complement, universe of discourse) and how we may use venn diagrams to help sort out the elements (or cardinalities of sets) of a universe of discourse.
Here, the universe of discourse ($U$) will be taken to be the integers from $100$ to $400$ (inclusive). $A$ will be taken to be the set those of elements of $U$ that are divisible by $2$; $B$, by $3$; $C$, by $5$; $D$, by $7$.
First, we will determine the cardinalities of (number of elements in) each of the sets $A,B,C,D$ (using the technique of common differences described by the author): $$|A|=151,\: |B|=100,\: |C|=61,\: |D|=43.$$ Next, we determine the cardinalities of the intersections of any pair of the sets $A,B,C,D$ (again with common differences): $$|A\cap B|=50,\: |A\cap C|=31,\: |A\cap D|=21,\: |B\cap C|=20,\: |B\cap D|=15,\: |C\cap D|=9.$$ Then we look at intersections of any trio of the sets $A,B,C,D$, similarly: $$|A\cap B\cap C|=10,\: |A\cap B\cap D|=7,\: |A\cap C\cap D|=4,\: |B\cap C\cap D|=3.$$ Now the intersection of all of them has only $210$ as an element, so $|A\cap B\cap C\cap D|=1$. Thus, there is only one number on the list divisible by all four of $2,3,5,7$. To find how many are divisible by exactly three, we take those divisible by at least three (in this case, members of intersections of trios) and toss out the one divisible by all four. In particular: $$|A\cap B\cap C\cap(\neg D)|=9,$$ $$|A\cap B\cap(\neg C)\cap D|=6,$$ $$|A\cap(\neg B)\cap C\cap D|=3,$$ $$|(\neg A)\cap B\cap C\cap D|=2,$$ so there are a total of $9+6+3+2=20$ numbers on the list divisible by exactly three of $2,3,5,7$.
To determine how many are divisible by exactly two, we take those divisible by at least two (members of intersections of pairs) and toss out those divisible by all four and those divisible by exactly three. For example, $|A\cap B\cap(\neg C)\cap(\neg D)|=50-1-9-6=34$, and we similarly find $$|A\cap(\neg B)\cap C\cap(\neg D)|=18,$$ $$|(\neg A)\cap B\cap C\cap(\neg D)|=8,$$ $$|A\cap(\neg B)\cap(\neg C)\cap D|=11,$$ $$|(\neg A)\cap B\cap(\neg C)\cap D|=6,$$ $$|(\neg A)\cap(\neg B)\cap C\cap D|=3,$$ giving us a total of $80$ numbers on the list divisible by exactly two of $2,3,5,7$.
For those divisible by exactly one, start with those divisible by at least one, and toss those divisible by more. So, $|A\cap(\neg B)\cap(\neg C)\cap(\neg D)|=151-1-6-3-9-34-18-11=69$, and simliarly, $$|(\neg A)\cap B\cap(\neg C)\cap(\neg D)|=34,$$ $$|(\neg A)\cap(\neg B)\cap C\cap(\neg D)|=17,$$ $$|(\neg A)\cap(\neg B)\cap(\neg C)\cap D|=11,$$ giving us $131$ numbers on the list divisible by exactly one of $2,3,5,7$. Unless I made a mistake, $131$ should be the answer to the second problem you encountered, if "either" and "any one of" are, in fact, intended to mean different things and "either" is a throwaway word. In total, then, there are $1+20+80+131=232$ numbers on the list that are divisible by at least one of $2,3,5,7$, and since there were only $301$ on the list in the first place, then there are $69$ that are divisible by none of $2,3,5,7$.
While time-consuming, the advantage of proceeding as above is that, at this point, you can answer any question along this line, since everything is nicely sorted out.