The notion of quasi-projective orbifold is generally accepted to contain at least the following: let $X$ be a (simply-connected) complex manifold, $G$ a group acting on $X$ by biholomorphisms, and containing a finite-index subgroup $G' \subset G$ acting freely on $X$ so that $X/G'$ is a quasi-projective variety. Then the orbifold $X/\!/G$ is quasi-projective.

If $G$ does not act effectively on $X$ (with $H \subset G$ the finite normal subgroup of elements which act trivially everywhere) it is "unlikely" that one can find a finite-index $G'$ acting freely. Suppose that $X/\!/(G/H)$ is a quasi-projective variety in the above sense: should $X/\!/G$ be considered quasi-projective?

Question 1 : If $G$ does not act effectively on $X$, can $X/\!/G$ be ever considered a quasi-projective orbifold in any sense?

In the above case, an algebraic line bundle on $X/\!/G$ can be taken to be a $G$-equivariant holomorphic line bundle on $X$ such that induced line bundle on $X/G'$ is algebraic.

Question 2 : What should one take as definition of algebraic line bundle if $X/\!/G$ is not effective?