Charles and Jonathan have given good answers to (1), but here's another way to recover the weak equivalences from the cofibrations and fibrations, which I think is sometimes convenient.

First of all, the wfss give us notions of "fibrant replacement" (a $(C_W,F)$-factorization of $X\to 1$) and "cofibrant replacement" (a $(C,F_W)$-factorization of $0\to X$). If $F_W\subseteq F$ (equivalently $C_W\subseteq C$), then we have two notions of fibrant-cofibrant replacement.

Now the wfs $(C_W,F)$ defines a notion of "path object", namely a $(C_W,F)$ factorization of the diagonal $X\to P X \to X\times X$. From this we can obtain a notion of "right homotopy": two maps $A\to X$ are right homotopic if the induced map into $X\times X$ factors through some path object for $X$. Dually, $(C,F_W)$ defines a notion of "cylinder object" and thereby "left homotopy".

If the two given wfss underlie a model structure, then the following all characterize the class of weak equivalences:

- $f\in W$ iff some (hence any) fibrant-cofibrant replacement of $f$ is a left (or right) homotopy equivalence (i.e. becomes an isomorphism upon quotienting by homotopy)
- $f\in W$ iff some (hence any) cofibrant replacement $f'$ of $f$ has the property that it induces an isomorphism on right homotopy classes of maps into any fibrant object.
- Dually, $f\in W$ iff some (hence any) fibrant replacement $f'$ of $f$ has the property that it induces an isomorphism on left homotopy classes of maps out of any cofibrant object.

In contrast to Charles' and Jonathan's answers, with these definitions it's almost automatic to get 2-out-of-3 (there's probably some fiddlyness with some vs any, but if you have a given functorial realization of your wfss you could just use that factorization to define the replacements). Moreover, $C_W$ automatically has the second property, while $F_W$ automatically has the third. Thus, if you can show that the three definitions agree, then you get $C_W \subseteq C\cap W$ and $F_W \subseteq F\cap W$. The tricky part would now be showing the reverse inclusions (although a standard argument implies that you only need to show one of them).