If a category is "monadic", is it necessarily so in a unique manner? Just a minor curiosity that's flitted across my mind, but that's (part of) what this site's for, right?:
Is it possible for Hom(a, -) and Hom(b, -) to both be monadic functors from C to Set, for non-isomorphic objects a and b in C? Ideally, the answer would come with either a nice example or an outline of a nice proof of impossibility (i.e., proof that all monadic representable functors on a category are isomorphic).
 A: You can certainly have non-equivalent monadic functors.  Here's one example: Let $\mathcal{V}_k$ be the category of $k$-vector spaces.  For a vector space $V$, let $H_V: \mathcal{V}_k\to \mathcal{V}_k$ be the functor
$$
H_V(W) = hom_k(V,W).
$$ 
Such a functor is always monadic, as long as $V$ is non-zero and finite dimensional.  The associated monad is
$$
T_V(W) = hom_k(V, V\otimes_k W) = End_k(V)\otimes_k W,
$$
so this is presenting a Morita equivalence: $k$-vector spaces are equivalent to modules over the matrix ring $End_k(V)$.
You wanted functors to set; let $U_V:\mathcal{V}_k\to Set$ be given by the same formula as $H_V$.  Then again, this will be monadic, as long as $V$ is non-zero and finite dimensional (and I'm not sure you even really need the finite dimensionality condition for either of these examples; added: you certainly don't in the first example, since $H_V$ is an exact functor, so the hypotheses of the Barr-Beck theorem certainly hold, though $T_V$ is not tensoring with an endomorphism ring if $V$ is infinite.).
