= Solution
A form of <Beck's monadicity theorem> compatible with equivalence-based <monadic adjunctions> is this. A <functor> $U:\mathcal D\to\mathcal C$ is monadic exactly when it has a <left adjoint>, is a <conservative functor>, and $\mathcal D$ has <coequalizers> of all pairs whose images under $U$ admit a <split coequalizer>, with $U$ preserving these coequalizers. In the strict lifting formulation, one states creation of these coequalizers. Such a pair is a <functor-split coequalizer pair>; the splitting is required in the base category, not necessarily upstairs.
A <split coequalizer> fork $A\mathrel{\substack{\xrightarrow{r}\\[-0.4ex]\xrightarrow[s]{}}}B\xrightarrow{q}Q$ has maps $u:B\to A$ and $t:Q\to B$ such that
$$
qr=qs,\qquad qt=1_Q,\qquad ru=1_B,\qquad su=tq.
$$
Every <functor> preserves this <coequalizer>: if $hr=hs$, then $h=hru=hsu=htq$, and $q$ is split epic, so the factor through $q$ is unique. The same equations remain true after applying any functor.
For necessity, let $U^T:\mathcal C^T\to\mathcal C$ be the <monad algebra> forgetful functor. It has the free-algebra <left adjoint>. It reflects <isomorphisms>, because if an algebra map $f$ is invertible in $\mathcal C$, the equation $fa=bTf$ rearranges to $f^{-1}b=aT(f^{-1})$.
If two algebra maps $r,s:(A,a)\rightrightarrows(B,b)$ have an underlying <split coequalizer> $q:B\to Q$, then $Tq$ and $T^2q$ are also coequalizers. Since $qb$ equalizes $Tr,Ts$, there is a unique $c:TQ\to Q$ satisfying
$$
cTq=qb.
$$
After composing with the <epimorphism> $q$, the equation $c\eta_Q=1_Q$ reduces to the unit law for $b$. After composing with the epimorphism $T^2q$, the equation $cT(c)=c\mu_Q$ reduces to $bT(b)=b\mu_B$. Thus $(Q,c)$ is an algebra and $q$ an algebra map. If an algebra map $h:(B,b)\to(Z,z)$ equalizes $r,s$, its unique underlying factor $k:Q\to Z$ satisfies $kc=zTk$: this follows by precomposing with the epimorphism $Tq$. Therefore this is a coequalizer upstairs as well. This proves that the <monad algebra forgetful functor creates split coequalizers>. The same properties transport through the comparison <equivalence of categories>, with the object-lifting convention distinguished in Question 9.
For sufficiency, suppose the three conditions hold, choose $F\dashv U$, and let $T=UF$ with comparison $K$. For each algebra $(X,a)$, consider the pair
$$
FTX\mathrel{\substack{\xrightarrow{F(a)}\\[-0.4ex]\xrightarrow[\varepsilon_{FX}]{}}}FX.
$$
Its image has coequalizer $a:TX\to X$. It is split: take $r=\mu_X$, $s=T(a)$, $q=a$, $u=\eta_{TX}$ and $t=\eta_X$. The <monad> and algebra laws give
$$
a\mu_X=aT(a),\quad a\eta_X=1_X,\quad
\mu_X\eta_{TX}=1_{TX},\quad T(a)\eta_{TX}=\eta_Xa.
$$
By hypothesis the pair has a coequalizer $q:FX\to D_X$ and $Uq$ is also a coequalizer. Identify $UD_X$ with $X$ using the unique coequalizer isomorphism, so $Uq=a$. Since $q$ is carried by $K$ to an algebra map from the free algebra, the transported action $b:TX\to X$ satisfies
$$
bT(a)=a\mu_X=aT(a).
$$
But $T(a)$ is a <split epimorphism>, with right inverse $T(\eta_X)$, so $b=a$. Hence $(X,a)\cong K(D_X)$: the comparison is essentially surjective.
Next, for every $D\in\mathcal D$, put $a_D=U\varepsilon_D$. The <counit of an adjunction> coequalizes
$$
FTUD\mathrel{\substack{\xrightarrow{F(a_D)}\\[-0.4ex]\xrightarrow[\varepsilon_{FUD}]{}}}FUD.
$$
This follows from naturality of the counit. The image pair has the split coequalizer $a_D$, as above. If $q:FUD\to Q$ is the hypothesized preserved coequalizer, the induced map $h:Q\to D$ has $Uh$ invertible because both $Uq$ and $a_D$ coequalize the same pair. Conservativity makes $h$ invertible. Therefore $\varepsilon_D$ itself is a coequalizer.
Let $f:K(D)\to K(E)$ be an algebra map, so $fa_D=a_ETf$. Set $g=\varepsilon_EF(f):FUD\to E$. Under the <adjunction> transpose bijection, the two composites of $g$ with the displayed presentation pair correspond respectively to $fa_D$ and $a_ETf$. They are equal, so $g$ factors uniquely through $\varepsilon_D$, giving $h:D\to E$ with $h\varepsilon_D=g$. Applying $U$ yields
$$
Uh\,a_D=a_ETf=fa_D.
$$
Since $a_D$ is split epic, $Uh=f$. Thus $K$ is full. If $Uh=Uk$, the transposes of $h\varepsilon_D$ and $k\varepsilon_D$ are equal, so those composites agree. The counit is epic, hence $h=k$; thus $K$ is faithful.
The comparison is full, faithful and essentially surjective, so it is an <equivalence of categories>. \b[This proves monadicity.] The argument also explains the role of each hypothesis: split base presentations construct all algebras, and reflection of isomorphisms turns the counit presentations into actual coequalizers upstairs.
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