= Solution
A <monad> on $\mathcal C$ is an endofunctor $T$ with natural transformations
$$
\eta:1_{\mathcal C}\Rightarrow T,
\qquad
\mu:T^2\Rightarrow T
$$
satisfying $\mu\,T\mu=\mu\,\mu T$ and $\mu\,T\eta=1_T=\mu\,\eta T$. Its <Eilenberg-Moore category> $\mathcal C^T$ has <algebras for a monad> $(A,a:TA\to A)$ satisfying
$$
a\eta_A=1_A,
\qquad
aT(a)=a\mu_A,
$$
and <morphisms of algebras for a monad> $f:(A,a)\to(B,b)$ satisfying $fa=bT(f)$.
The <Kleisli category> $\mathcal C_T$ has the objects of $\mathcal C$ and
$$
\mathcal C_T(A,B)=\mathcal C(A,TB).
$$
Its identity is $\eta_A$, while the composite of $f:A\to TB$ and $g:B\to TC$ is $\mu_C T(g)f$. The <free functor into a Kleisli category> $F_T:\mathcal C\to\mathcal C_T$ is the identity on objects and sends $f:A\to B$ to $\eta_Bf$.
On the <functor category> $[\mathcal D,\mathcal C]$, postcomposition gives the <pointwise monad on a functor category>
$$
T_*(F)=TF,
\qquad
(\eta_*)_F=\eta F,
\qquad
(\mu_*)_F=\mu F.
$$
The monad laws hold componentwise. A $T_*$-algebra is a functor $F:\mathcal D\to\mathcal C$ with a natural transformation $a:TF\to F$ whose components are $T$-algebras. Naturality says precisely that every $F(u)$ is an algebra morphism. Hence sending $(F,a)$ to the lifted functor $\mathcal D\to\mathcal C^T$ gives an isomorphism, and in particular an equivalence,
$$
[\mathcal D,\mathcal C]^{T_*}\simeq[\mathcal D,\mathcal C^T].
$$
On $[\mathcal C,\mathcal D]$, precomposition gives the <precomposition monad on a functor category>
$$
T^*(G)=GT,
\qquad
(\eta^*)_G=G\eta,
\qquad
(\mu^*)_G=G\mu.
$$
A $T^*$-algebra is a natural transformation $a:GT\to G$ satisfying $aG\eta=1_G$ and $a(aT)=aG\mu$. From it define $\bar G:\mathcal C_T\to\mathcal D$ by
$$
\bar G(A)=G(A),
\qquad
\bar G(f:A\to TB)=a_BG(f).
$$
The two algebra laws say exactly that $\bar G$ preserves identities and Kleisli composition, and $\bar G F_T=G$. Conversely, a factorization $G=\bar G F_T$ gives
$$
a_A=\bar G(1_{TA}:TA\to TA),
$$
where $1_{TA}$ represents a Kleisli arrow $TA\to A$. These constructions are inverse on objects and morphisms, so
$$
[\mathcal C,\mathcal D]^{T^*}\simeq[\mathcal C_T,\mathcal D].
$$
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