The Wikipedia article you linked to describes the different types of completeness fairly well, but there are some interesting connections between them. First, semantic completeness (everything true is provable) does not imply syntactic completeness (each sentence or its negation is provable), since there are sentences which are contingent (e.g., $A \lor B$), and the propositional calculus (a semantically complete calculus) won't be able to prove either of $A \lor B$ or $\lnot(A \lor B)$.
To the first question, a deductively complete system need not be consistent. The trivial counterexample is that an inconsistent system proves everything, and thus proves $\phi$ or $\lnot\phi$ for every $\phi$ (in fact, it proves both!).
To the second question, a deductively complete system need not be (semantically) complete. We could construct, for instance, an esoteric calculus with bizarre proof rules that guarantee that for every formula $\phi$, either $\phi$ or $\lnot\phi$ is a theorem. For instance, given the propositional language with variables $A$, $B$, $C$, $\dots$, the connectives $\lnot$, $\land$, $\lor$, and $\to$, where well-formed formulae are defined as usual, take as your system just the following two axiom schemata:
- $\phi$ where $\phi$ is a wff in which $\land$ appears an odd number of times
- $\lnot\phi$ where $\phi$ is a wff in which $\land$ does not appear an odd number of times
Since every formula $\phi$ either has an odd number of $\land$s, in which case $\phi$ is a theorem, or it doesn't, in which case $\lnot\phi$ is a theorem, the system is deductively complete. Theorems of this system include
\begin{gather*}
\lnot A \qquad A \land B \qquad \lnot(A \land (B \land C))
\end{gather*}
It's pretty obviously not semantically complete (there are true sentences that it does not prove) and even unsound (there are non-true sentences that it does prove), but it's syntactically consistent (it never proves $\phi$ and $\lnot\phi$).
To sum up, a deductively complete system need not be (syntactically) consistent (never prove $\phi$ and $\lnot\phi$), sound ($\vdash \phi$ implies $\models \phi$), or (semantically) complete ($\models \phi$ implies $\vdash\phi$).
It would be an interesting exercise to determine which combinations of those can be had in a proof system, though. In the examples above saw deductively complete systems that were:
- inconsistent, unsound, and (semantically) complete;
- consistent, unsound, and (semantically) incomplete;
Without specifying a particular interpretation for the propositional variables, we cannot have a sound deductively complete system, because the deductively complete system has to make some judgment about the contingent sentences. Any inconsistent system will automatically be (semantically) complete, so the only possible case that we haven't seen is one which is:
- consistent, unsound, and (semantically) complete.
Such a system could be had if a propositional variable assignment were part of the proof system. For instance, take the standard propositional calculus as a starting point. Then suppose some particular interpretation $\cal I$ of propositional variables (as an easy one, let every propositional variable be true). The standard propositional calculus is consistent and semantically complete, so we have those. To get the unsoundness, let us also add the axiom schema inference rule:
- $\phi$, if $\cal I \models \phi$.
This will have the effect of making every otherwise contingent sentence (or its negation) a theorem. Since the propositional calculus is semantically complete, the only $\phi$ such that neither $\phi$ nor $\lnot\phi$ is a theorem are contingent sentences, but now those have been fixed as well, so the resulting system is consistent, unsound (since, e.g., $\vdash A \land B$ and $\not\models A \land B$), and semantically complete.
