(*I completely revamped my answer, the previous version ([link](http://mathoverflow.net/revisions/163925/2)) had a consistency result with a particular example, this feels much better as it establishes a full equivalence result instead.*)

The answer is no. For projectivity and injectivity. The reason is that the axiom of choice is equivalent to the following statement:

> If $V$ is a vector space, and $K$ is a subspace of $V$, then there exists a a direct complement for $K$, that is a subspace $S$ such that $S\cap K=\{0\}$ and $S\cup K$ is a generating set for $V$.

If every vector space is projective, consider the case that $V$ is a vector space, $K$ is a subspace and $T\colon V\to V/K$ is the quotient map. Then there is some $h\colon V/K\to V$ such that $h\circ T$ is the identity function. In particular, this means that $K+\operatorname{im}(h)=V$, but $h$ cannot map any nonzero vector into $K$ so this is a direct sum indeed.

Therefore the statement "Every vector space is projective" implies the axiom of choice.

Similarly for injectivity, take the identity function $K\to V$, then there is a projection map whose kernel is a direct complement for $K$.

----

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>