I do not know an efficient way to determine whether a quiver representation is projective or injective. The definitions and properties such as "Projectives are summands of free modules", etc do not seem to be easy to work with.
For example does the quiver representation $$0\rightarrow0\leftarrow k\rightarrow k\leftarrow k\leftarrow k\rightarrow k\rightarrow k\leftarrow 0$$ have a projective cover? Thank you very much!
If $k$ is a field, and your quiver has no oriented cycles, then there are exactly as many indecomposable projective modules (up to isomorphism) as there are vertices in the quiver. Moreover, any finite-dimensional representation admits a projective cover. (This is all true because of more general results on Artinian algebras).
In your specific example, to find the projective cover of the representation, try to find its top, and then compute its projective cover. Here, this should be $P_3\oplus P_6$, where $P_i$ is the projective module "at vertex $i$" (meaning that if $A$ is the path algebra, and $e_i$ is the path of length zero at vertex $i$, then $P_i = A e_i$).