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No. Its top nonzero group would force dimension three. Poincare duality requires $H_2\cong H^1$, and the universal coefficient theorem gives $H^1\cong \mathrm{Hom}(H_1,\mathbb{Z})=\mathbb{Z}$, since $H_0$ is free. This contradicts the proposed vanishing of $H_2$. Equivalently its listed Betti numbers give Euler characteristic $-1$, impossible for a closed oriented odd-dimensional manifold.

No. The dimension would again be three. The universal coefficient theorem gives $H^1=0$, since both $\mathrm{Hom}(H_1,\mathbb{Z})$ and $\mathrm{Ext}(H_0,\mathbb{Z})$ vanish. Poincare duality then forces $H_2=0$, contradicting the proposed torsion group. In general the second integral homology of a connected closed oriented three-manifold is free abelian, being dual to its first homology.

The universal coefficient theorem relates homology or cohomology with different coefficient groups through tensor and Tor, or Hom and Ext, respectively. The universal coefficient theorem for homology is its tensor-Tor sequence; the universal coefficient theorem for cohomology is its Hom-Ext sequence.

This is the cohomological form of the universal coefficient theorem, with Ext and Hom terms rather than the homological tensor and Tor terms.

For an abelian coefficient group $G$, cohomology fits into a split short exact sequence

For an abelian coefficient group $G$, there is a split short exact sequenceFor $G=\mathbb{Q}$, the Tor term vanishes.

This is the homology case of the universal coefficient theorem; its cohomology case uses Hom and Ext rather than tensor product and Tor.