= Solution
The <terminal object> $1$ is an <exponentiable object> because $-\times1\cong1_{\mathcal C}$ has the identity as a <right adjoint>. If $A$ and $B$ are <exponentiable objects>, compose the two <adjunctions> to get
$$
\begin{aligned}
\mathcal C(X\times(A\times B),Y)
&\cong\mathcal C((X\times A)\times B,Y)\\
&\cong\mathcal C(X\times A,Y^B)
\cong\mathcal C(X,(Y^B)^A).
\end{aligned}
$$
Consequently \b[<exponentiable objects> are closed under finite <products in a category>], including the empty product, and
$$
\boxed{Y^{A\times B}\cong(Y^B)^A.}
$$
Under the additional hypotheses, a <coseparator> $S$ detects unequal parallel arrows by postcomposition: for $u\ne v:X\to B$ there is $t:B\to S$ with $tu\ne tv$. Local smallness makes $I=\mathcal C(B,S)$ a set, and completeness forms the <product in a category> $S^I$. The evaluation arrow
$$
e_B:B\to S^I,\qquad\pi_t e_B=t
$$
is a <monomorphism>, because $e_Bu=e_Bv$ forces $tu=tv$ for every $t$, and the <coseparator> then forces $u=v$.
By hypothesis this <monomorphism> is a <regular monomorphism>, so it is an <equalizer> of some $r,s:S^I\rightrightarrows Q$. Apply the same evaluation construction to $Q$: $J=\mathcal C(Q,S)$ is a set and $e_Q:Q\to S^J$ is monic. Cancelling $e_Q$ shows that equalizing $e_Qr,e_Qs$ is equivalent to equalizing $r,s$. We have therefore proved the <equalizer presentation by powers of a coseparator>:
$$
\boxed{B\xrightarrow{e_B}S^I\mathrel{\substack{\xrightarrow{e_Qr}\\[-2pt]\xrightarrow[e_Qs]{} }}S^J.}
$$
The notation $S^I$ here means a product indexed by a set, so its existence does not presuppose categorical exponentiation.
If $A$ is an <exponentiable object>, $\mathcal C(-\times A,S)\cong\mathcal C(-,S^A)$ is a <representable functor>. Conversely, suppose a representing object $E$ and a natural isomorphism
$$
\mathcal C(X,E)\cong\mathcal C(X\times A,S)
$$
are given. For any set $I$, the <product in a category> property gives
$$
\mathcal C(X,E^I)\cong\mathcal C(X\times A,S^I).
$$
Thus maps into all powers of $S$ have representations. For an <equalizer> presentation $B\to S^I\rightrightarrows S^J$ with arrows $r,s$, postcomposition induces <natural transformations> between these represented functors. The <Yoneda lemma> identifies them with arrows $r_A,s_A:E^I\rightrightarrows E^J$. Form their <equalizer> $B_A$. Then
$$
\begin{aligned}
\mathcal C(X,B_A)
&\cong\operatorname{Eq}\bigl(\mathcal C(X,E^I)\rightrightarrows\mathcal C(X,E^J)\bigr)\\
&\cong\operatorname{Eq}\bigl(\mathcal C(X\times A,S^I)\rightrightarrows\mathcal C(X\times A,S^J)\bigr)\\
&\cong\mathcal C(X\times A,B).
\end{aligned}
$$
These bijections are natural in $X$. For a morphism $B\to C$, postcomposition and the <Yoneda lemma> give the unique arrow $B_A\to C_A$ respecting them; uniqueness proves functoriality. Hence $B\mapsto B_A$ is a <right adjoint> to $-\times A$. This proves the <exponentiability test using a coseparator>:
$$
\boxed{A\text{ is exponentiable}\iff\mathcal C(-\times A,S)\text{ is representable}.}
$$
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