$\text{Ext}(H, \mathbb{Z})$ is isomorphic to the torsion subgroup of $H$ if $H$ is finitely generated

The third property (with $G=\mathbb{Z}$) tells you that if $C$ is a finite cyclic group, then $\operatorname{Ext}(C,\mathbb{Z})\cong C$. Furthermore, any finitely generated torsion abelian group is a direct sum of cyclic groups. So $T$ is a direct sum of cyclic groups, so by the first and third properties, $\operatorname{Ext}(T,\mathbb{Z})\cong T$. Note, however, that the isomorphism in the third property is not canonical, so while $\operatorname{Ext}(H,\mathbb{Z})$ is isomorphic to $T$, it is not canonically isomorphic to $T$.

To elaborate on my comment, applying $\text{Ext}^{\bullet}(A, -)$ to the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$ produces the long exact sequence

$$0 \to \text{Hom}(A, \mathbb{Z}) \to \text{Hom}(A, \mathbb{R}) \to \text{Hom}(A, S^1) \to \text{Ext}^1(A, \mathbb{Z}) \to 0$$

where we can ignore the rest of the sequence because $\text{Ext}^1(A, \mathbb{R})$ vanishes, thanks to the fact that $\mathbb{R}$ is divisible and hence injective. (So is $S^1$; in fact what we've written down above is an injective resolution of $\mathbb{Z}$.)

If $A$ is torsion, then $\text{Hom}(A, \mathbb{R}) = 0$, and the long exact sequence above produces a natural isomorphism

$$\text{Hom}(A, S^1) \cong \text{Ext}^1(A, \mathbb{Z}).$$

The group $\text{Hom}(A, S^1)$ is known as the Pontryagin dual $\widehat{A}$ of $A$. If $A$ is finite it is noncanonically isomorphic to $A$. In general it naturally has a topology making it a profinite abelian group, and every profinite abelian group arises in this way.