= Solution
Put $P(A)=\Omega^A$, the <power object>, with $P(f)$ acting by inverse image of predicates. Transposing a predicate on $A\times B$ gives natural bijections
$$
\mathcal E(A,PB)\cong\operatorname{Sub}(A\times B)\cong\mathcal E(B,PA).
$$
Thus $P^{\mathrm{op}}:\mathcal E\to\mathcal E^{\mathrm{op}}$ is left adjoint to $P:\mathcal E^{\mathrm{op}}\to\mathcal E$. Its unit $\eta_A:A\to PPA$ sends $a$ to evaluation at $a$, namely the predicate $S\mapsto(a\in S)$. This unit is monic: two generalized elements having the same evaluation on all predicates agree by testing the singleton predicate of the first, or equivalently its graph in the relevant parameter product.
We verify a sufficient form of <Beck's monadicity theorem>: a right adjoint is monadic if it reflects <isomorphisms> and its domain has, and it preserves, <coequalizers> of reflexive pairs. Suppose $Pf$ is invertible. Then $PPf$ is invertible, and naturality of the monic unit shows that $f$ is monic. Regard it as the <subobject> $A\hookrightarrow B$. Its characteristic predicate and the constantly true predicate have the same restriction to $A$, so injectivity of $Pf$ makes them equal. Thus this <subobject> is all of $B$, and $f$ is invertible. Hence $P$ reflects <isomorphisms>.
A reflexive pair in $\mathcal E^{\mathrm{op}}$ is a coreflexive pair $f,g:B\rightrightarrows A$ in $\mathcal E$, with a common retraction $r$ such that $rf=rg=1_B$. Its <equalizer> $e:E\hookrightarrow B$ exists. We show that $Pe:PB\to PE$ is the <coequalizer> of $Pf,Pg:PA\rightrightarrows PB$. Direct image of <subobjects> along the mono $f$ defines $\exists_f:PB\to PA$, and similarly $\exists_e:PE\to PB$. At every parameter object the identities are
$$
Pf\,\exists_f=1_{PB},\qquad Pg\,\exists_f=\exists_ePe,\qquad Pe\,\exists_e=1_{PE}.
$$
For the middle identity, an equality $f(b)=g(b')$ forces $b=b'$ after applying $r$; their common point must therefore lie in the <equalizer>. This proves the identity for parameterized predicates, not merely for global elements. If $h:PB\to Z$ satisfies $hPf=hPg$, composing with $\exists_f$ gives $h=h\exists_ePe$. Thus it factors through $Pe$, and the last displayed identity gives uniqueness. This proves that <power objects turn coreflexive equalizers into coequalizers>. All hypotheses of the monadicity criterion have now been checked, so \b[$P$ is monadic], with
$$
\boxed{\mathcal E^{\mathrm{op}}\simeq\mathcal E^{PP}.}
$$
A <logical functor> $L:\mathcal E\to\mathcal F$ preserves <finite limits>, exponentials and the classifier, hence gives compatible natural <isomorphisms> $LP\cong PL$ and $LPP\cong PPL$. They respect units and multiplications, because these are defined by evaluation and transposition. Under the displayed equivalences, $L^{\mathrm{op}}$ is therefore the induced functor on double-power <Eilenberg-Moore categories>. Their forgetful functors create <finite limits>, and the underlying $L$ preserves them. Consequently the induced functor preserves <finite limits>, so $L^{\mathrm{op}}$ does too. Reversing arrows proves \b[every <logical functor> preserves finite <colimits>].
For the first counterexample, the constant-terminal functor $F:\mathbf{Set}\to\mathbf{Set}$ has the constant-empty functor as left adjoint: both sets of maps in the <adjunction> are singletons. It preserves exponentials, since $F(B^A)=1=1^1=(FB)^{FA}$ with the canonical comparison. It does not preserve the <initial object>, because $F(0)=1$. Thus preservation of exponentials alone is insufficient.
For the second, take the fixed-point functor $\Gamma:G\text{-}\mathbf{Set}\to\mathbf{Set}$ for $G=\mathbb Z/2$. Its left adjoint gives a set the trivial action. The classifier of $G$-sets is the two-element set with trivial action, so $\Gamma$ preserves it and its truth map. Let $X$ be the free two-element $G$-set. The <coequalizer> of its identity and its swapping automorphism is the singleton $G$-set. Applying $\Gamma$ gives two maps $0\rightrightarrows0$ followed by $1$, whereas their <coequalizer> in sets is $0$. Thus \b[$\Gamma$ preserves the classifier but not all finite <colimits>]. These examples also explain why both requirements in logicalness matter.
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