Cole, my answers agree with yours. Assuming that an adjoint exists, proposition 1.88 gives a way to construct it, so I used that, simplified a bit and then verified that the functions I got were the required adjoints.

I think in the case of Puzzles 12 and 13 the adjoints are unique. Suppose \$$f : A \rightarrow B \$$ is left adjoint to \$$g \$$ and that we can uniquely determine an element \$$x \$$ of \$$A \$$ by the set of elements \$$\\{ a \in A | a \leq x \\} \$$. Then the choice of \$$f \$$ seems to uniquely determine the value of \$$g(b) \$$ for every \$$b \in B \$$, since
$$a \leq g(b) \Leftrightarrow f(a) \leq b$$

EDIT: To clarify, suppose \$$d : A \rightarrow \mathcal P (A) \$$ defined by
\$$d(x) = \\{ a \in A | a \leq x \\} \$$
has a left inverse \$$h \$$. Then
\$$g(b) = h(d(g(b))) = h( \\{ a \in A | a \leq g(b) \\} ) = h( \\{ a \in A | f(a) \leq b \\} ) \$$