Solutions

1.

The answer is ``No." S_g is

$$\sim P M (\sim P M)$$

2.

Proof. S_g is true if and only if the mirror of $\sim P M$ is not printable. But the mirror of $\sim P M$ is S_g itself. Hence S_g is true iff it's not printable. By the definition of the biconditional, we thus have two possibilities: viz., either S_g is true and not printable, or S_g is printable and false. The second possibility violates one of the assumptions of the puzzle (viz., that $\cal M$₀ never prints sentences that aren't true). So we are left with the other disjunct: S_g is true but not printable. QED

3.

As to $P M (\sim P M)$, it must be false (since it's negation is true). Therefore by the hypotheses governing the puzzle this sentence is not printable. If we reinterpret `printable' as `provable' we can conclude that $P M (\sim P M)$ is, as we say, undecidable, that is, neither it nor its negation is provable.