The transition function is holomorphic at $\infty$ so the line bundle is trivial.

$\mathrm{PSL}_2(R)$ acts on the circle seen as the real projective line and the universal cover of $\mathrm{PSL}_2(R)$ (which is the same as the universal cover of $\mathrm{SL}_2(R)$) acts on the universal cover of the circle, i.e., the real line. Hence it is not clear that this is a good example.

The Hodge index theorem gives you that $K^2_SC^2\le(C\cdot K_S)^2$.

The complement has no quotient on which $B$ acts trivially, which means that it is spanned by elements of the form $ba-a$ which are commutators in $G$ so the complement is contained in $G'$. On the other hand $G$ modulo the complement is clearly commutative so that $G'$ is contained in it.

You don't say what your context is (ringed spaces, schemes,...). If it is schemes then no, as the stalk of the product is a local ring and the tensor product of local may not be local.

BTW, the Riemann existence is in Forster (Cor. 14.13 in my edition).

Yes, it is immediate, we have $e^2=e$ and $be=e$. The first shows that $A$ is the direct sum of the image and the kernel of $e$, the second that the image lies in $C$ and and it is clear that $e$ is the identity on $C$.

I don't know what you would consider a oneliner but the usual proof in the context of ordinary representation theory works. We get an action of $e=1/|B|\sum_{b\in B}b$ on $A$ and it gives the projection on $C$.

$D^\times$-torsors over some $X$-scheme $Y$ correspond to locally free (formally local in the flat topology) $D_Y$-modules of rank $1$. Such modules are locally free already in the Zeriski topology. Now, the obstruction to lifting a section of $i_\ast F$ is such a torsor and hence the section is liftable Zariski-locally.