Answer to Question #1
Paraconsistent mathematical theories will not always be complete. Depending on what the theory takes to be true, and the strength of its deductive system, there might well be unprovable truths. As I'm sure you are by now aware, all paraconsistent theories give up ex falso quodlibet (the rule that allows you to derive anything from a contradiction) as well as principles that entail it (like disjunctive syllogism: from "A or B" and "not-B" deduce "A"). This means that inconsistencies within these theories will not "explode" allowing the proof of any statement of the language. Thus, paraconsistency is no guarantee of completeness. Embracing inconsistency does, however, open the door to the possibility of a complete theory whose classical counterpart would be essentially incomplete. For a toy example, a paraconsistent theory which keeps ex falso quodlibet (though, such a theory wouldn't really be paraconsistent anymore) as an admissible inference will be trivially complete (I imagine this is something like what you had in mind).
Answer to Question #2
Well, many interesting paraconsistent theories won't be consistent, so those theories certainly shouldn't be able to prove their own consistency--- that would be bad. I'm not quite sure what else you had in mind, but it is interesting to note that Tarski's corollary of Gödel's results--- Tarski's Undefinability Theorem ---is no longer much threat. If you look at the linked Shapiro article (in "Further Reading") you'll see that the theory he develops is a paraconsistent arithmetic (a dialetheist arithmetic to be more precise; I suspect that many of your questions about paraconsistent theories are really meant to be about dialetheist or otherwise inconsistent theories) which contains its own truth predicate. It can prove its own soundness and its own Gödel sentence.
SEP Article on Inconsistent Mathematics
IEP Article on Inconsistent Mathematics
Incompleteness and Inconsistency; by Stewart Shapiro