Is $\Bbb S^2 \times \Bbb S^4$ symplectic?

There is no symplectic form on $\mathbb{S}^2\times\mathbb{S}^4$. More generally, there is no symplectic form on $M\times \mathbb{S}^{2n}$, if $n>1$ and $M$ is compact, see Symplectic structures on $M\times \mathbb{S}^{2n}$ .

No. Note that $H^2(S^2\times S^4,\mathbb R)$ is one dimensional, spanned by $\pi^*\alpha$, where $\pi:S^2\times S^4\to S^2$ is the projection, and $\alpha$ is a volume form on $S^2$. Suppose $\omega$ is a symplectic form on $S^2\times S^4$. Then $[\omega]=c[\pi^*\alpha]$ for some $c\in\mathbb R^\times$. Then $[\omega^3]=c^3[\pi^*\alpha^3]=0$, contradicting the requirement that $\omega^3$ is everywhere nondegenerate.