= Solution
Write a <coalgebra for a comonad> as $(A,a)$ with $a:A\to GA$, $\epsilon_Aa=1_A$ and $\delta_Aa=Ga\,a$. Since the <Cartesian comonad> $G$ preserves <finite limits>, the <forgetful functor> $U:\mathcal E_G\to\mathcal E$ creates them: for an underlying limit $K$ of coalgebras, the maps $K\to A_i\to GA_i$ induce a unique $K\to GK$ using $GK\cong\lim_iGA_i$. Its counit and coassociativity equations follow after every limiting projection, which jointly detect equality. The same argument makes the limiting cone universal among coalgebra cones. Thus $\mathcal E_G$ is a <Cartesian category>; in particular, the product structure on $A\times B$ is $a\times b$ followed by the comparison $GA\times GB\cong G(A\times B)$.
An <exponential object> needs more than the underlying ambient $B^A$. The <cofree coalgebra> functor is $R(D)=(GD,\delta_D)$, with <adjunction>
$$
\mathcal E_G((C,c),R(D))\cong\mathcal E(C,D),\qquad
h\longmapsto\widetilde h=Gh\,c.
$$
The inverse sends $\widetilde h$ to $\epsilon_D\widetilde h$; the <comonad> laws verify that these are inverse and that $\widetilde h$ is a coalgebra map.
For $(A,a)$ and $(B,b)$, put $D=B^A$ and $W=(GB)^A$ in the ambient <Cartesian closed category>. Define $p,q:GD\to W$ by
$$
p=b^A\epsilon_D,
$$
and by transposing the composite
$$
GD\times A\xrightarrow{1\times a}GD\times GA
\cong G(D\times A)\xrightarrow{G\operatorname{ev}}GB
$$
to obtain $q$. Their cofree-adjoint transposes are coalgebra maps
$$
p^\sharp=Gp\,\delta_D,\qquad q^\sharp=Gq\,\delta_D:
R(D)\rightrightarrows R(W).
$$
Take their coalgebra <equalizer> $e:H\hookrightarrow R(D)$, which exists by the finite-limit construction.
To verify its universal property, let $h:C\to D$ transpose to $t:C\times A\to B$. Its coalgebra transpose $\widetilde h=Gh\,c$ factors through $H$ exactly when $p^\sharp\widetilde h=q^\sharp\widetilde h$. By the cofree <adjunction>, this equality is equivalent to $p\widetilde h=q\widetilde h$. The first side is $b^Ah$, because $\epsilon_DGh\,c=h$; transposing the second side gives $Gt$ composed with the product coalgebra structure on $C\times A$. The equality therefore says exactly
$$
bt=Gt\,(c\times a),
$$
with the canonical product comparison understood: precisely the condition that $t$ is a coalgebra map. Thus
$$
\boxed{\mathcal E_G(C,H)\cong\mathcal E_G(C\times A,B)}
$$
naturally in the coalgebra $C$. Its evaluation is ambient evaluation after $\epsilon_De\times1_A$, and the preceding condition makes it a coalgebra map. \b[This equalizer is the coalgebra exponential, so the category of coalgebras is cartesian closed.] This <Cartesian closure for a finite-limit-preserving comonad> uses no subobject classifier on $\mathcal E$.
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