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As for question (1): assuming cocomplete cartesian monoidal category means that cartesian products distribute over colimits (and I know from past experience that you do mean this, Martin), then it's true that the monad $T$ preserves reflexive coequalizers. The main thing you need is that finitary power functors $c \mapsto c^n$ preserve reflexive coequalizers. This is a corollary of a result that you can find on the first page of chapter 0 of Johnstone's Topos Theory, which is a $3 \times 3$ lemma stating that if the rows and columns are reflexive coequalizer diagrams, then so is the diagonal. It easily follows from this lemma that for example the squaring functor $c \mapsto c \times c$ preserves reflexive coequalizers, and a similar inductive argument allows you to extend this to any finite power $c \mapsto c^n$.

To derive the fact that monads $T: C \to C$ based on a Lawvere algebraic theory $\theta$ preserve reflexive coequalizers, write

$$T(c) = \int^{n \in \theta} mathrm{FinSet}^{op}} \hom_\theta(n, 1hom_\theta(i(n), i(1)) \cdot c^{|n|}$$ c^n$$where the tensor S \cdot c of a set S with an object c is the coproduct of copies of c indexed over S. (Here i: \mathrm{FinSet}^{op} \to \theta denotes the unique (up to isomorphism) map of Lawvere algebraic theories, viewing \mathrm{FinSet}^{op} as the "initial" Lawvere algebraic theory.) Since coend functors and tensor functors S \cdot - preserve reflexive coequalizers, as does c \mapsto c^{|n|}c^{n}, we see that T does as well. I can't think of a more direct nice description of the composite left adjoint F, nor do I think one is needed because I think the description you gave is plenty nice. As for (2): the underlying functor is definitely not monadic. It's not even a right adjoint, because for example for C = \mathrm{Vect}_k, it fails to preserve the terminal object (which in \mathrm{AbHopf}(\mathrm{Vect})_k is the monoidal unit k, as is the case just in \mathrm{CoMon}(\mathrm{Vect})_k). Edit: Since this came up in comments, let me provide an alternative proof of the fact that finite power functors on a cocomplete cartesian monoidal category preserve reflexive coequalizers. Recall that a category J is sifted if the diagonal functor J \to J \times J is final (result due to Gabriel and Ulmer). A prototypical example is where J is the generic parallel pair equipped with a section in common. Then follow Steve Lack's soft proof here, which uses just the assumption that C is cocomplete cartesian monoidal and the finality of the diagonal on J, to show the binary product C^2 \to C preserves reflexive coequalizers. Similarly, the n-fold product C^n \to C preserves reflexive coequalizers. The n-fold power on C is a composite of the diagonal \Delta: C \to C^n (which is a left adjoint, thus colimit-preserving) with the n-fold product, so it too preserves reflexive coequalizers. 2 added 1100 characters in body Edit: Since this came up in comments, let me provide an alternative proof of the fact that finite power functors on a cocomplete cartesian monoidal category preserve reflexive coequalizers. Recall that a category J is sifted if the diagonal functor J \to J \times J is final (result due to Gabriel and Ulmer). A prototypical example is where J is the generic parallel pair equipped with a section in common. Then follow Steve Lack's soft proof here, which uses just the assumption that C is cocomplete cartesian monoidal and the finality of the diagonal on J, to show the binary product C^2 \to C preserves reflexive coequalizers. Similarly, the n-fold product C^n \to C preserves reflexive coequalizers. The n-fold power on C is a composite of the diagonal \Delta: C \to C^n (which is a left adjoint, thus colimit-preserving) with the n-fold product, so it too preserves reflexive coequalizers. 1 As for question (1): assuming cocomplete cartesian monoidal category means that cartesian products distribute over colimits (and I know from past experience that you do mean this, Martin), then it's true that the monad T preserves reflexive coequalizers. The main thing you need is that finitary power functors c \mapsto c^n preserve reflexive coequalizers. This is a corollary of a result that you can find on the first page of chapter 0 of Johnstone's Topos Theory, which is a 3 \times 3 lemma stating that if the rows and columns are reflexive coequalizer diagrams, then so is the diagonal. It easily follows from this lemma that for example the squaring functor c \mapsto c \times c preserves reflexive coequalizers, and a similar inductive argument allows you to extend this to any finite power c \mapsto c^n. To derive the fact that monads T: C \to C based on a Lawvere algebraic theory \theta preserve reflexive coequalizers, write$$T(c) = \int^{n \in \theta} \hom_\theta(n, 1) \cdot c^{|n|}

where the tensor $S \cdot c$ of a set $S$ with an object $c$ is the coproduct of copies of $c$ indexed over $S$. Since coend functors and tensor functors $S \cdot -$ preserve reflexive coequalizers, as does $c \mapsto c^{|n|}$, we see that $T$ does as well.

I can't think of a more direct nice description of the composite left adjoint $F$, nor do I think one is needed because I think the description you gave is plenty nice.

As for (2): the underlying functor is definitely not monadic. It's not even a right adjoint, because for example for $C = \mathrm{Vect}_k$, it fails to preserve the terminal object (which in $\mathrm{AbHopf}(\mathrm{Vect})_k$ is the monoidal unit $k$, as is the case just in $\mathrm{CoMon}(\mathrm{Vect})_k$).