= Solution
For an <adjoint functor> pair $F:\mathcal D\rightleftarrows\mathcal C:G$, the <unit and counit of an adjunction> are
$$
\eta:1_{\mathcal D}\Rightarrow GF,
\qquad
\varepsilon:FG\Rightarrow1_{\mathcal C}.
$$
Under the adjunction bijection, $\eta_A$ corresponds to $1_{FA}$ and $\varepsilon_B$ corresponds to $1_{GB}$. They satisfy the triangle identities
$$
\varepsilon_{FA}F(\eta_A)=1_{FA},
\qquad
G(\varepsilon_B)\eta_{GB}=1_{GB}.
$$
Conversely, natural transformations with these identities recover the adjunction through the mutually inverse maps
$$
f:FA\to B\longmapsto G(f)\eta_A,
\qquad
g:A\to GB\longmapsto\varepsilon_BF(g).
$$
The <fully faithful adjoint criterion> gives the first equivalence directly. If $F$ is <full and faithful>, there is a unique $r_A:GFA\to A$ with $F(r_A)=\varepsilon_{FA}$; the triangle identity and faithfulness show that $r_A$ and $\eta_A$ are inverse, so the unit is an isomorphism. If the unit is an isomorphism, the displayed adjunction bijection shows that
$$
\mathcal D(A,A')\longrightarrow\mathcal C(FA,FA')
$$
is bijective, so $F$ is full and faithful. This also proves that either condition gives a natural isomorphism $GF\cong1_{\mathcal D}$.
For the remaining direction, suppose merely that $GF$ is naturally isomorphic to the identity. Transport the <monad induced by an adjunction> along this isomorphism. Its underlying endofunctor is then the identity, its unit is a natural endomorphism $u:1\Rightarrow1$, and its multiplication is a natural endomorphism $m:1\Rightarrow1$ with $mu=1$. Naturality makes $u_A$ commute with $m_A$, so also $um=1$. Hence the transported unit, and therefore $\eta$, is an isomorphism. The three conditions are equivalent.
Now let $F\dashv G\dashv H$. If $F$ is full and faithful, then $GF\cong1_{\mathcal D}$. For $A,B\in\mathcal D$, the two adjunctions give natural bijections
$$
\mathcal D(B,GHA)
\cong\mathcal C(FB,HA)
\cong\mathcal D(GFB,A)
\cong\mathcal D(B,A).
$$
The <Yoneda lemma> therefore gives $GHA\cong A$, naturally in $A$. The <fully faithful adjoint criterion> applied to $G\dashv H$ shows that $H$ is full and faithful. Conversely, if $H$ is full and faithful, then $GH\cong1_{\mathcal D}$ and
$$
\mathcal D(GFA,B)
\cong\mathcal C(FA,HB)
\cong\mathcal D(A,GHB)
\cong\mathcal D(A,B).
$$
Another application of the <Yoneda lemma> gives $GFA\cong A$, so $F$ is full and faithful. Thus $F$ is full and faithful exactly when $H$ is.
Assume henceforth that $G$ is full and faithful. Then the counit $\varepsilon:FG\to1_{\mathcal C}$ and unit $\alpha:1_{\mathcal C}\to HG$ are natural isomorphisms. Consider
$$
\theta_1=(\alpha F)^{-1}(H\eta),
\qquad
\theta_2=(F\beta)(\varepsilon H)^{-1}:H\Rightarrow F.
$$
Applying the faithful functor $G$, then using naturality and the four triangle identities, turns both composites into
$$
GH\xrightarrow{\beta}1_{\mathcal D}\xrightarrow{\eta}GF.
$$
Therefore $\theta_1=\theta_2$; denote their common value by $\theta$, the <double-adjoint comparison transformation>.
The pointwise monicity criterion is clearest from the following natural square, in which both vertical maps are bijections:
$$
\begin{array}{ccc}
\mathcal C(B,HA)&\xrightarrow{\theta_A\circ-}&\mathcal C(B,FA)\\
\downarrow&&\downarrow\\
\mathcal D(GB,A)&\xrightarrow{F}&\mathcal C(FGB,FA).
\end{array}
$$
The left vertical map is the adjunction $G\dashv H$, while the right one precomposes with the isomorphism $\varepsilon_B:FGB\to B$. Thus every $\theta_A$ is a <monomorphism> exactly when $F$ is faithful on all morphisms $GB\to A$, namely morphisms whose domains lie in the image of $G$.
Dually, the natural square
$$
\begin{array}{ccc}
\mathcal C(FA,B)&\xrightarrow{-\circ\theta_A}&\mathcal C(HA,B)\\
\downarrow&&\downarrow\\
\mathcal D(A,GB)&\xrightarrow{H}&\mathcal C(HA,HGB)
\end{array}
$$
has bijective vertical maps, using $F\dashv G$ on the left and the isomorphism $\alpha_B:B\to HGB$ on the right. Hence every $\theta_A$ is an <epimorphism> exactly when $H$ is faithful on all morphisms $A\to GB$, namely morphisms whose codomains lie in the image of $G$.
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