Given a functor $$F$$, a monad $$M=(F, \mu,\eta)$$ and a comonad $$N=(F,\delta, \xi)$$, do $$M, N$$ always compose to form a bimonad?
Not necessarily. Take for example any pointed category $$\mathcal{C}$$, and let $$1$$ denote the category with exactly one morphism. There is a unique functor $$\mathcal{C}\to 1$$, and the functor $$1\to \mathcal{C}$$ that just picks the zero object is both its left and right adjoint. So composing the two gives an endofunctor of $$\mathcal{C}$$ that just maps every object to the zero object of $$\mathcal{C}$$, and this functor has a monad and a comonad structure.
However they cannot form a bimonad structure in the sense of this paper. Indeed one of the conditions there is that the composition of the unit and counit is the identity transformation of the identity endofunctor on $$\mathcal{C}$$; but here the unit is $$X\to 0$$ and the counit is $$0\to X$$, so their composition is the zero transformation and not the identity.