= Solution
For $T=GF$, define the <Eilenberg-Moore comparison functor>
$$
\boxed{K(D)=(GD,a_D),\qquad a_D=G\varepsilon_D,\qquad K(h)=Gh.}
$$
The algebra unit law is a triangle identity. Naturality of the counit at $\varepsilon_D$ gives its algebra associativity law; naturality at $h:D\to E$ gives $Gh\,a_D=a_E T(Gh)$, making $K(h)$ an algebra morphism.
Suppose first that $G$ reflects G-split <coequalizers>. For every $D$, consider the counit presentation
$$
FGFGD\underset{FG\varepsilon_D}{\overset{\varepsilon_{FGD}}{\rightrightarrows}}FGD\xrightarrow{\varepsilon_D}D.
$$
It equalizes the pair by counit naturality. After applying $G$ it becomes $T^2GD\rightrightarrows TGD\xrightarrow{a_D}GD$, with parallel maps $\mu_{GD},T(a_D)$. This is split: take $s=\eta_{GD}$ and $t=\eta_{TGD}$. The identities $a_Ds=1$, $\mu_{GD}t=1$ and $T(a_D)t=\eta_{GD}a_D$ follow from the algebra law, a <monad> unit law, and naturality of $\eta$. Reflection therefore makes $\varepsilon_D$ a <coequalizer> in $\mathcal D$.
For faithfulness, if $Gh=Gk$ then naturality gives $h\varepsilon_D=\varepsilon_EFGh=\varepsilon_EFGk=k\varepsilon_D$. <Coequalizers> are epic, so $h=k$.
For fullness, let $\alpha:GD\to GE$ be an algebra morphism, so $\alpha a_D=a_E T\alpha$. Put $h_0=\varepsilon_EF\alpha:FGD\to E$. Its two composites with the counit-presentation pair are equal: by counit naturality they are $\varepsilon_EF(a_ET\alpha)$ and $\varepsilon_EF(\alpha a_D)$. Hence there is a unique $h:D\to E$ with $h\varepsilon_D=h_0$. Applying $G$ gives $Gh\,a_D=a_ET\alpha=\alpha a_D$. Since $a_D$ is split epic, $Gh=\alpha$. This proves full faithfulness directly by descent.
Conversely suppose $K$ is full and faithful. We need the following explicit algebra lifting fact. For algebra morphisms $f,g:(A,a)\rightrightarrows(B,b)$ with a split underlying <coequalizer> $q:B\to Q$, the arrow $qb:TB\to Q$ equalizes $Tf,Tg$. The split diagram is preserved by $T$, so $Tq$ is their <coequalizer> and there is a unique $c:TQ\to Q$ with $cTq=qb$. Precomposing the unit equation with the <epimorphism> $q$ gives $c\eta_Qq=q$; precomposing the associativity equation with the <epimorphism> $T^2q$ gives equality by $bT(b)=b\mu_B$. Thus $c$ is an algebra action. If an algebra map $h:B\to(Z,z)$ equalizes $f,g$, its unique underlying factor $k:Q\to Z$ satisfies $kcTq=hb=zTh=zTkTq$. Cancelling $Tq$ shows $kc=zTk$. This proves <monad algebra forgetful functor creates split coequalizers>.
Now take any existing equalizing arrow $e:B\to D$ in $\mathcal D$ whose image is such a <coequalizer>. The lifting fact makes the unique compatible action on $GD$ its already given action $a_D$, since $Ke$ preserves that action. Hence $Ke$ is a <coequalizer> of $Kf,Kg$ in the algebra category. For any arrow $h:B\to Z$ equalizing the pair, its unique algebra factor $KD\to KZ$ lifts to an arrow $D\to Z$ by fullness of $K$; faithfulness gives its factorization equation and uniqueness. Therefore $e$ is a <coequalizer> in $\mathcal D$. We have proved the <full comparison and reflection of split coequalizers> equivalence
$$
\boxed{K\text{ is full and faithful}\iff G\text{ reflects G-split coequalizers}.}
$$
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