Question consists of three parts.
Prove that a symplectic Torus admits no Hamiltonian circle actions
Prove that a symplectic 4_Torus admits no Hamiltonian Torus actions
Consider a circle action on $T^2 x S^2$ where $\theta \in S^1$ acts on a pair $(\theta_1,\theta_2) \in T^2$ and $(r,s,z) \in S^2$ (in cylindric coordinates) by sending it to $(\theta_1+n\theta , \theta_2+m\theta)$ and $(r,s+\theta,z)$. For which n and m it is a Hamiltonian action?
I have already part 1. We know that any action on Torus has not fixed points iff it is not Hamiltonian. (By McDuff theorem) But I could not solve part 2 and 3.
To answer 2, By a result of Hui Li https://arxiv.org/abs/math/0605133, if $M$ is a symplectic manifold with a Hamiltonian $S^{1}$-action then $\pi_{1}(M) \cong \pi_{1}(M_{\min})$, where $M_{\min}$ is the subset where the Hamiltonian attains its minimum. $M_{\min}$ is a connected, symplectic submanifold of $M$.
If $M$ is a 4-torus, it cannot have a Hamiltonian circle action. Assume it did for a contradiction. Clearly $\dim(M_{\min}) \leq 2$, but there is no connected, symplectic manifold of dimension at most two with fundamental group $\mathbb{Z}^{4}$, which is a contradiction to the above result.