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 \\} ) \\)

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 \\} ) \\)