= Solution
Assume $F$ is a <full and faithful functor> and its essential image is closed under <strong quotients>. First its unit is invertible. To prove this directly, fullness gives $k:GFA\to A$ with $Fk=\varepsilon_{FA}$. The triangle identity and faithfulness give $k\eta_A=1_A$. <Naturality> of $\eta$ at $k$ and the other triangle identity give
$$
\eta_Ak=GFk\,\eta_{GFA}=G\varepsilon_{FA}\,\eta_{GFA}=1_{GFA}.
$$
Thus $\eta_A$ is an <isomorphism>, and $\varepsilon_{FA}=(F\eta_A)^{-1}$ is also an <isomorphism>.
Factor $\varepsilon_B$ as a <strong epimorphism> followed by a <monomorphism>. By closure under <strong quotients>, the intermediate object can, after transport along an <isomorphism>, be written as $FA$:
$$
FGB\xrightarrow{e}FA\xrightarrow{m}B,\qquad \varepsilon_B=me.
$$
Fullness gives $e=Fk$ for $k:GB\to A$. Taking the transpose of $\varepsilon_B=mFk$ under the <adjunction>, whose transpose is $1_{GB}$, gives
$$
1_{GB}=Gm\,GFk\,\eta_{GB}=Gm\,\eta_A\,k.
$$
Hence $k$ is a <split monomorphism>, so $e=Fk$ is a <monomorphism>. A <strong epimorphism> which is monic is invertible: apply its lifting property to the square with that same map on both sides and identity top and bottom edges. Thus $e$ is invertible and $\varepsilon_B=me$ is monic. We have proved the <pointwise-monic unit-and-counit criterion>:
$$
\boxed{\eta,\varepsilon\text{ monic}\iff F\text{ fully faithful with image closed under strong quotients}.}
$$
For a counterexample without the balanced hypothesis, use the <pointwise-monic adjunction over a non-balanced poset>. Let $\mathcal C=\{0<1\}$ and let $\mathcal D$ have one object and only its identity. The unique $F:\mathcal C\to\mathcal D$ is left adjoint to $G$ selecting $1$, since both $\mathcal D(Fc,*)$ and $\mathcal C(c,1)$ are singletons. All <morphisms> in a <poset> viewed as a <category> are <monomorphisms>, so the unit and counit are monic. However $F$ is not full: the identity $F1\to F0$ has no preimage $1\to0$. The arrow $0\to1$ is both monic and epic but not invertible, so $\mathcal C$ is not a <balanced category>. \b[The balanced hypothesis cannot be dropped.]
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