Both answers are "No."

There are well-known obstructions to the existence of an equivariant momentum mapping arising from the action by symplectomorphisms of a group $G$ on a symplectic manifold. They can be phrased in many ways, but if $G$ is connected and its Lie algebra is semisimple, for example, the obstructions vanish.

A nice treatment can be found in the classic paper by Atiyah and Bott: "The moment map and equivariant cohomology" in Topology 1984.

**After-dinner update:**

Let me try to add some more details, since it seems I was a little too quick both reading the question and also the comments! (In my defense, I was getting really hungry and wanted to get home!)

Let $(M,\omega)$ be a symplectic manifold and let $G$ be a connected Lie group acting on $M$ via symplectomorphisms. Let $\mathfrak{g}$ be the Lie algebra of $G$ and for every $X \in \mathfrak{g}$ let $\xi_X$ denote the corresponding vector field on $M$. Since $G$ acts symplectomorphically, $\xi_X$ is a *symplectic vector field*; that is, $\mathcal{L}_{\xi_X} \omega = 0$, which, as Ben points out, implies that $d i_{\xi_X}\omega = 0$. Let $\mathrm{Sym}(M)$ denote the Lie algebra of symplectic vector fields on $M$. We have a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$.

A symplectic vector field $\xi$ is said to be *hamiltonian* if $i_\xi\omega$ is not merely closed, but also exact. The $G$-action is *hamiltonian* if the $\xi_X$ are hamiltonian for all $X \in \mathfrak{g}$. Let $\mathrm{Ham}(M)$ denote the Lie algebra of hamiltonian vector fields. It is not hard to show that the Lie bracket of two symplectic vector fields is hamiltonian, so $\mathrm{Ham}(M)$ is an ideal in $\mathrm{Sym}(M)$ with abelian quotient. This gives rise to a short exact sequence of Lie algebras
$$
0 \longrightarrow \mathrm{Ham}(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0
$$
where $H^1(M)$ is the first de Rham cohomology group thought of as an abelian Lie algebra.

If $\xi \in \mathrm{Ham}(M)$, then $i_\xi \omega = df$ for some $f \in C^\infty(M)$. This defines a map $C^\infty(M) \to \mathrm{Ham}(M)$ which is also a Lie algebra homomorphism if we make $C^\infty(M)$ into a Lie algebra via the Poisson bracket. The kernel of this map consists of the locally constant functions, whence we have another short exact sequence of Lie algebras
$$
0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Ham}(M) \longrightarrow 0
$$

Putting these two sequences together we have a four-term exact sequence starting and ending at the first two de Rham cohomology groups:
$$
0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0
$$

Now the symplectic $G$-action defines a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$ and for the existence of a momentum map, we want this to lift to a Lie algebra morphism $\mathfrak{g} \to C^\infty(M)$.

There is an immediate obstruction for the map to lift at the level of vector spaces, namely the map $\mathfrak{g} \to \mathrm{Sym}(M) \to H^1(M)$ is a Lie algebra cocycle with values in the trivial module $H^1(M)$, and so defines a class in $H^1(\mathfrak{g};H^1(M))$. If this class vanishes, we do get a map $\mathfrak{g} \to C^\infty(M)$ which may only be a homomorphism modulo $H^0(M)$. In other words, it defines a Lie algebra homomorphism to a central extension of $C^\infty(M)$ defined by a 2-cocycle with values in the trivial module $H^0(M)$. The moment map will exist if the class of this cocycle in $H^2(\mathfrak{g}; H^0(M))$ vanishes.

For example, if $\mathfrak{g}$ is semisimple, then both $H^1$ and $H^2$ vanish and the momentum map exists.

The beautiful observation of Atiyah and Bott is that the obstruction can be reinterpreted in terms of the Cartan model for the equivariant de Rham cohomology of $M$, where it becomes simply the obstruction to extending $\omega$ to an equivariant cocycle.