I think everything is explained by the fact that your exact sequence is backwards. There is an exact sequence $0\rightarrow I_p\rightarrow \mathcal{O}\rightarrow T\rightarrow 0$ which when you tensor with $\Omega_C$ gives the result.

Haha, my friend and I barely managed to work through the logic with the audience choosing the numbers 2 and 3!

Thanks for the reference, this is exactly what I was looking for.

One direct way to show $\kappa(S)=-\infty$ implies $S$ is ruled is to use Iitaka's Conjecture: If $S$ is a surface and $S\rightarrow B$ is a fibration with generic fiber $F$ then $\kappa(S)\geq\kappa(B)+\kappa(F)$. Applying this to the Albanese fibration solves case (2) above instantaneously, because it implies that $F$ must be rational. This applies the stronger assumption $P_n=0$ for all $n$ rather than $P_{12}=0$ though. I think there really will be no way to get the specific number $12$ without some classification.

Thanks for your answer, it partially resolves the case of my second question, by giving criteria for contractibility in the algebraic category in certain cases. I mainly wanted an explicit construction {\it in the analytic category} of the contraction. (The conditions would of course be weaker if we allow the contraction not to be an algebraic surface). Since the proposition above seems to be the best result about contractibility, I assume it is hard then to determine whether the resulting surface is algebraic...

This works whether or not $D$ is torsion, for any $(1,1)$-class in $H^2(X,\mathbb{Z})$.