= Solution
A <balanced category> is one in which every morphism that is both monic and epic is an <isomorphism>. If a <faithful functor> sends $f$ to an <isomorphism>, then $f$ is monic: equality $fg=fh$ becomes $FfFg=FfFh$, cancellation gives $Fg=Fh$, and faithfulness gives $g=h$. The same argument on the other side makes $f$ epic. Balance therefore makes $f$ an <isomorphism>. Thus a <faithful functor> out of a <balanced category> reflects <isomorphisms>.
Under the <adjunction>, the transpose of $Ff:FX\to FA$ is $\eta_A f:X\to GFA$. Hence if $F$ is faithful, $\eta_Af=\eta_Ag$ forces $Ff=Fg$ and then $f=g$, making every $\eta_A$ monic. Conversely, if the unit is pointwise monic and $Ff=Fg$, naturality gives $\eta_Bf=GFf\,\eta_A=GFg\,\eta_A=\eta_Bg$, whence $f=g$. Therefore
$$
\boxed{F\text{ faithful}\iff\eta\text{ pointwise monic}.}
$$
We also use the standard <adjunction> fact with its justification: $F$ is full and faithful exactly when $\eta$ is invertible. Indeed the composite hom-set map $\mathcal C(X,A)\to\mathcal D(FX,FA)\cong\mathcal C(X,GFA)$ is postcomposition by $\eta_A$. If $F$ is full and faithful these maps are bijective; taking $X=GFA$ produces a right inverse, and taking $X=A$ gives injectivity and therefore a left inverse. Conversely an invertible unit makes these hom-set maps bijective.
For the final example, view the two-element ordered set $0<1$ as a <category> $\mathcal C$, and let $\mathcal D$ have one object and only its identity. The unique $F:\mathcal C\to\mathcal D$ is left adjoint to the <functor> $G$ selecting $1$, since every $\mathcal C(x,1)$ is a singleton. Every arrow in an ordered-set <category> is monic, so the unit arrows $x\to1$ and the identity counit are monic. But $F$ is not full: $\mathcal C(1,0)$ is empty whereas $\mathcal D(F1,F0)$ is a singleton. The example does not contradict the upcoming equivalence because $\mathcal C$ is not balanced; its arrow $0\to1$ is both monic and epic but not invertible.
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