Suppose you have a representation of $x_1\lor x_2\lor x_3$ as 2CNF clauses, possibly involving with hidden variables. (This would "represent" the three-way disjunction in the sense that the 2CNF is satisfiable together with any consistent combination of $(\neg)x_i$ that contains at least one positive $x_i$, and is not satisfiable together with $\neg x_1 \land \neg x_2 \land \neg x_3$.
First any hidden variable $h$ can be eliminated by replacing all clauses involving $h$ or $\neg h$, by the result of applying resolution to every combination of such clauses that has $h$ in one and $\neg h$ in another. This does not change the satisfiability of the formula. (Here an important observation is that using resolution on two 2-clauses produces another 2-clause; the is not true for clauses of higher arity).
Thus, without loss of generality we can assume that there are no hidden variables -- and the question is then just whether any 2CNF in the variables $x_1, x_2, x_3$ is logically equivalent to $x_1\lor x_2\lor x_3$.
We can easily see that this is not the case. Any nontrivial 2-clause will be true for at most $6$ of the $8$ different truth valuations of $x_1, x_2, x_3$. Taking the conjunction of several such clauses cannot make more valuations come out true -- but $x_1\lor x_2\lor x_3$ is true in $7$ cases, so it is not equivalent to any 2CNF.
I assume that for a) the task is to come up with a formula that says that exactly one of the variables is true.
This can be done fairly easily in DNF:
$$(x_1 \land \neg x_2 \land \neg x_3 \land \neg x_4 \land \neg x_5) \lor (\neg x_1 \land x_2 \land \neg x_3 \land \neg x_4 \land \neg x_5) \lor ...$$
(You can do the rest ... obviously you'll get $5$ terms like this)
However, how do we express this in CNF, as b) asks?
Well, you can always take a formula in DNF, and transform it into CNF by Distribution of $\lor$ over $\land$
For example:
$$(P \land Q) \lor (R \land S) \Leftrightarrow (P \lor R) \land (P \lor S) \land (Q \lor R) \land (Q \lor S)$$
(This should be reminiscent of the FOIL principle)
In your case, however, you'd get $5^5=3125$ terms .... and while many of them can be eliminated by further logical principles, this is not something you'd actually want to do ...
OK, can we do something else?
Yes, consider all possible ways in which it would be false that exactly one of the variables is true. So that would be when they are either all false, or when two or more of them are true, i.e.:
$$(\neg x_1 \land \neg x_2 \land \neg x_3 \land \neg x_4 \land \neg x_5) \lor (x_1 \land x_2) \lor (x_1 \land x_3) \lor ... \lor (x_1 \land x_5) \lor (x_2 \land x_3) \lor ...$$
(again, you can do the rest ... you'll get $1 + {5 \choose 2}=11$ terms)
OK, so since this statement is the negation of the statement you want, simply negate this statement to get the statement you want. And note: by doing a bunch of DeMorgan's, you'll get a statement nicely in CNF!
Best Answer
The point of 3CNF is that it is a "normal form" for formulas: as you mention, every formula is equivalent, up to a quadratic (linear?) blowup, to a 3CNF formula. 3CNF formulas are "simple", and so more easy to deal with. In particular, if you ever read about NP-completeness, you will find out that we want our to put our "challenging problems" in as simple a form as possible. This makes it easier both to design and analyze algorithms solving these problems, and to prove that other problems are also difficult by reducing 3CNF to them (showing how to solve 3CNF using an algorithm for them).
We care specifically about 3CNF for historical reasons, it was the first (or one of the first) NP-complete problems (check out Cook's paper or Karp's paper on their respective pages). Also, 2CNF is not "complete" (arbitrary formulas cannot are not equivalent to a 2CNF), and it is easy to determine whether they are satisfiable or not (google if interested).
The conversion from CNF to 3CNF is best explained by an example. We convert each clause separately. The clause $A \lor B \lor C \lor D \lor E$ is equivalent to the 3CNF $$(A \lor B \lor x_1) \land (\lnot x_1 \lor C \lor x_2) \land (\lnot x_2 \lor D \lor E), $$ in the sense that the original formula (in this case, a single clause) is satisfiable iff the converted formula is. You convert each of the clauses, and take their conjunction.
Suppose you have an arbitrary formula using the connectives $\lnot,\lor,\land$. Picture it as a tree, where inner nodes are labeled with $\lnot,\lor,\land$, and each node has either one ($\lnot$) or two ($\lor,\land$) children. I'm sorry I can't provide a picture. The first step is "pushing the negations to the leaves" using de Morgan's rules (q.v.). That rids us of all $\lnot$ nodes, but now leaves may be literals (variables or negations of variables). Now we convert the formula to a CNF recursively.
Denote by $\varphi$ the function we're constructing, that takes a formula as above (with $\land,\lor$ and possibly negated variables) and returns a CNF. For the base case, we have $\varphi(x) = x$ and $\varphi(\lnot x) = \lnot x$. For a formula of the form $A \land B$, we don't have to work hard: we define $\varphi(A \land B) = \varphi(A) \land \varphi(B)$. The hardest case is formulas of the form $A \lor B$: a somewhat economical choice is $$\varphi(A \lor B) = (y \lor \varphi(A)) \land ((\lnot y) \lor \varphi(B)), $$ where $y$ is a new variable, and $y \lor \varphi(A)$ means adding $y$ to all the clauses.
General formulas with arbitrary connectives can be handled by expressing the connectives using $\lor,\land,\lnot$, for example by writing a truth-table; this can be quite wasteful, though.
The entire construction results in a quadratic blowup in formula size (your "occurrences of variables"; there are many other ways to define formula size). However, the result you're quoting is only a linear blowup (indeed, by a factor of $24$). It is entirely possible that such a construction exists, but I'm not aware of it; perhaps one of the readers is. The SAT to 3SAT part has a linear blowup with a factor of $3$.
If you do not allow adding new variables, then no simple conversion is possible. While it is always possible to convert an arbitrary formula into a CNF (using the "truth table" approach), not every formula is convertible to a 3CNF without adding new variables. An example is parity on $4$ variables. Also, the conversion from an arbitrary formula to a CNF can result in exponential blow-up, for example for parity on $n$ variables we go from $\Theta(n^2)$ to $\Theta(n2^n)$.