Not really: you can't get all divisors that way. That's a fascinating story you can search for in courses and textbooks on something like "algebraic geometry" or "geometry of surfaces".

That should be $X\cap H$ above.

Every time you have $X$ in projective space, you have a divisor $X\cup H$ on $X$ where $H$ is a hyperplane... Take a look at any book on projective varieties.

Thanks! I misread the question as being about finitely presented modules. As Robin Chapman correctly says, the question is stated needs some additional assumptions!

I don't think you'll get anything this way that you couldn't get by Verma modules... although that, again, depends on what types of constructions you're looking for. Perhaps you should really split the question into different ones...