= Solution
The two introductory requests can be settled before the lettered applications. In the <functor category> $[C,\mathbf{Set}]$, <finite limits> and finite unions of <subobjects> are pointwise. The component of the diagonal at $c$ is ordinary equality on $F(c)$. Its only possible complement is
$$
D(c)=\{(a,b)\in F(c)^2\mid a\ne b\}.
$$
This forms a subfunctor exactly when each $F(u)$ sends unequal elements to unequal elements, equivalently when every $F(u)$ is injective. In that case $D$ and the diagonal are disjoint and their union is $F^2$ at every component. Conversely, a diagonal complement must have these components and be stable under every transition map. Hence \b[$F$ is decidable exactly when every transition map is injective].
Regard a <monoid> $M$ as a one-object category; a covariant set-valued <functor> is a left <M-set>. Give $M\times A$ the diagonal left action $t\cdot(w,a)=(tw,t\cdot a)$. Let
$$
E=\operatorname{Hom}_M(M\times A,B),\qquad
(m\cdot f)(w,a)=f(wm,a).
$$
Right multiplication on the first coordinate commutes with the diagonal left action, so $m\cdot f$ remains equivariant. The formula obeys $m\cdot(n\cdot f)=(mn)\cdot f$ and $1\cdot f=f$. Evaluation is
$$
\operatorname{ev}:E\times A\to B,\qquad \operatorname{ev}(f,a)=f(1,a).
$$
It is equivariant because $\operatorname{ev}(m\cdot f,m\cdot a)=f(m,m\cdot a)=m\cdot f(1,a)$.
For an equivariant $h:C\times A\to B$, define
$$
\widehat h(c)(w,a)=h(w\cdot c,a).
$$
This is equivariant in $(w,a)$, and $\widehat h(m\cdot c)=m\cdot\widehat h(c)$. Evaluation recovers $h$. Conversely, currying the evaluation of a map $C\to E$ recovers that map by its equivariance. This proves the exponential universal property and the natural identification
$$
\boxed{B^A\cong\operatorname{Hom}_M(M\times A,B)}
$$
with exactly the stated action. In particular, <decidability in a set-valued functor category> says that a left <M-set> is decidable if and only if each of its action maps is injective.
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