Here's a basic answer to question 1: what makes a collection of sets a proper class?  The answer has to do with models, so let's look at those for a moment.

A model of set theory, also called a "universe", is a collection $M$ of things we'll call "sets", but for now they don't have any extra structure, they're just points or objects.  But a model also has some extra information in its definition, a relation $E\subset M\times M$.  And I'll write the relation as an infix, so that $xEy$ is short for $(x,y)\in E$.  This relation is how the model describes when one set is supposed to be an element of another: $xEy$ is supposed to satisfy all the axioms of set theory.  That means $M$ and $E$ only make a model of set theory if statements like
> For all $x,y\in M$, if $zEx$ holds whenever $zEy$ holds, then $x=y$. (Extensionality)

are true about $M$ and $E$.

Now, models of set theory don't have to "feel" like a universe of sets.  For example, you can construct the "set" of real numbers in any model of set theory, and prove that the set of real numbers is uncountable.  Well then, the model had better be uncountable for the real numbers to fit inside, right? No, in fact the [Löwenheim-Skolem theorem][1] implies that there's a model of set theory where $M$, the collection of all "sets", is countable.  (Assuming there's any model at all, of course.)  

Isn't that a contradiction? If $M$ is countable, then $r$ has a countable number of elements, so we can put them in one-to-one correspondence with the elements of $n$ (the "set" of natural numbers), and use Cantor's diagonal argument to find an element not on the list, etc.

Let's see why there actually isn't a contradiction:  If $r\in M$ is the "set" of real numbers and $n\in M$ is the set of natural numbers, we can consider the collection of all "elements" of $r$, namely $R=\{x\in M: xEr\}$, and the collection of "elements" of $n$, namely $N=\{x\in M: xEn\}$ Then if $M$ is a countable model, $R$ and $N$ are both countable, so we can find a bijection $N\to R$... But there the reasoning stops.  In order to get a contradiction with Cantor's diagonal argument, what we actually need there to be is a "set" $f\in M$ such that $f$ satisfies the conditions to be a  "function" from $n$ to $r$, but that isn't what we have.  

That leads us to the notion of "class" and "proper class".  If we're working with a model $M$, then sets are elements of $M$ and "classes" of sets are subsets of $M$.  Now some classes correspond to sets: any set $x\in M$ gives us a class of its elements: `X=$\{y\in M:yE x\}$.  Since the axiom of extensionality means that a set is determined by its class of elements, it's natural to identify a set with its class of elements.  In this way, we can ask whether a class $C\subset M$ corresponds to some set $c\in M$.  If it does, i.e. $x\in C$ iff $x E c$, we can abuse terminology and call $C$ a set, but if there's no such $c$, we call $C$ a proper class.

As the other answers have pointed out, there are many reasons $C$ can fail to be a set and so be a proper class. $C$ could be too "big", like $M$ itself: $M$ never corresponds to a set $m\in M$ by Russel's paradox: we'd have $xE m$ iff $x\in M$ which is true, so $m$ would be a "set" of all "sets".  Or $C$ could be a set that would give you something impossible if it existed, like a bijection between the set $r\in M$ of real numbers and $n\in M$ of natural numbers.  Or $C$ could just be a subset of $M$ that's outside the scope of $M$ for no good reason, in which case there are standard ways of building a new model $M'\supset M$ in which $C\subset M\subset M'$ actually does correspond to an "set" $c\in M'$.

This is one way to look at how independence results are shown: you start with a model $M$ of set theory, and then build a new model $M'$ that is not only a model of set theory but also has other properties, such as every "subset" of "the real numbers" being "in bijection" either with "the real numbers" or with "the natural numbers", so that $M'$ satisfies the continuum hypothesis.  Then you can rephrase that result as "If ZFC is consistent (has a model $M$), then so is ZFC+CH (there's a model $M'$ of both)."

  [1]: http://en.wikipedia.org/wiki/Lowenheim%25E2%2580%2593Skolem_theorem