= Solution
Define the <free algebra functor> by
$$
F^T(X)=(TX,\mu_X),\qquad F^T(f)=Tf.
$$
The <monad> laws say precisely that $\mu_X$ defines an <monad algebra> structure on $TX$, while <naturality> of $\mu$ makes $Tf$ an <monad algebra> <morphism>. We construct the <adjunction> explicitly.
For an <monad algebra> $(A,a)$ and a <morphism> $f:X\to A$, put $\overline f=aTf:TX\to A$. It is an <monad algebra> <morphism> because
$$
\overline f\mu_X=aTf\mu_X=a\mu_A T^2f
=aTaT^2f=aT\overline f.
$$
Conversely, from an <monad algebra> <morphism> $h:(TX,\mu_X)\to(A,a)$ obtain $h\eta_X:X\to A$. These operations are inverse: <naturality> of $\eta$ and the <monad algebra> unit law give
$$
aTf\eta_X=a\eta_A f=f,
$$
and the algebra-morphism equation together with the <monad> unit law gives
$$
aT(h\eta_X)=aThT\eta_X=h\mu_XT\eta_X=h.
$$
The formulas commute with precomposition in $X$ and composition with <monad algebra> <morphisms> in $(A,a)$, so the <bijections> are natural. Therefore the <free-forgetful Eilenberg-Moore adjunction> is
$$
\boxed{F^T\dashv U,\qquad\mathcal C^T(F^TX,(A,a))\cong\mathcal C(X,A).}
$$
Its unit is $\eta_X$ and its counit at $(A,a)$ is the <monad algebra> action $a$.
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