The answer is no. Or at least not necessarily. At least for projectivity.

It is consistent that there is a vector space $V$ over $\Bbb F_2$ (or really any finite field) whose underlying set is amorphous, namely it cannot be partitioned into two infinite subsets.

This vector space is not finitely generated, but every proper subspace has a finite dimension.

Pick any such subspace $V'$ and consider $W=V/V'$. Then it is not very difficult to show that the following holds:

1. $W$ is a vector space whose underlying set is amorphous, and every proper subspace is finitely generated. But $|W|\neq|V|$.
2. There is no non-zero homomorphism from $W$ to $V$.
3. The natural quotient map is surjective from $V$ to $W$.

So $W$ is not projective. This also gives an example of an exact sequence which doesn't split. As Jeremy Rickard observes in the comment, this is also a counterexample to injectivity.

(The above can be done easily over infinite fields too, but then $V$ is not amorphous, just weird, and we might not have $|W|\neq|V|$ in general.

Another generalization is that we might allow subspaces to be not just finitely generated, but generated by a well-ordered set up to some cardinal $\lambda$ (which would mean the subspace is free, of course), but the space itself is not generated by any such set.)

I think that a relevant paper to mention here would be the following:

> <cite cite="_Trans. Amer. Math. Soc._ **255** (1979), 31--59" mrnumber="542870" authors="Andreas Blass">_Andreas Blass_, [**Injectivity, projectivity, and the axiom of choice**](http://dx.doi.org/10.2307/1998165), _Trans. Amer. Math. Soc._ **255** (1979), 31--59.</cite>