Since $\mathbf{Q}/\mathbf{Z} = \varinjlim (1/n) \mathbf{Z}/\mathbf{Z}$ and $V = \varinjlim V[n]$ (as $V$ is torsion), it suffices to show that for $n > 0$ the natural map ${\rm{Hom}}(\mathbf{Q}/\mathbf{Z},V)/(n) \rightarrow V[n]$ defined by evaluation at $1/n \bmod \mathbf{Z}$ is an isomorphism (as then with a small diagram chase we can pass to the direct limit over more divisible $n$ to conclude).

But $V$ is an injective abelian group (as $V$ is divisible), so applying ${\rm{Hom}}(\cdot, V)$ to the exact sequence $$0 \rightarrow (1/n)\mathbf{Z}/\mathbf{Z} \rightarrow \mathbf{Q}/\mathbf{Z} \stackrel{n}{\rightarrow} \mathbf{Q}/\mathbf{Z} \rightarrow 0$$ yields exactly the desired isomorphism.

[In effect, this is a reformulation of the same argument as in Neil Strickland's answer, as I am masking some calculations implicit in the proof that divisible abeian groups are injective. I am also tacitly using that $\widehat{\mathbf{Z}}$ is the endomorphism ring of $\mathbf{Q}/\mathbf{Z}$, so tensoring over it against a torsion module is the same as tensoring over $\mathbf{Z}$.]

Since $\mathbf{Q}/\mathbf{Z} = \varinjlim (1/n) \mathbf{Z}/\mathbf{Z}$ and $V = \varinjlim V[n]$ (as $V$ is torsion), it suffices to show that for $n > 0$ the natural map ${\rm{Hom}}(\mathbf{Q}/\mathbf{Z},V)/(n) \rightarrow V[n]$ defined by evaluation at $1/n \bmod \mathbf{Z}$ is an isomorphism (as then with a small diagram chase we can pass to the direct limit over more divisible $n$ to conclude).

But $V$ is an injective abelian group (as $V$ is divisible), so applying ${\rm{Hom}}(\cdot, V)$ to the exact sequence $$0 \rightarrow (1/n)\mathbf{Z}/\mathbf{Z} \rightarrow \mathbf{Q}/\mathbf{Z} \stackrel{n}{\rightarrow} \mathbf{Q}/\mathbf{Z} \rightarrow 0$$ yields exactly the desired isomorphism.

[In effect, this is a reformulation of the same argument as in Neil Strickland's answer, as I am masking some calculations implicit in the proof that divisible abelian groups are injective. I am also tacitly using that $\widehat{\mathbf{Z}}$ is the endomorphism ring of $\mathbf{Q}/\mathbf{Z}$, so tensoring over it against a torsion module is the same as tensoring over $\mathbf{Z}$.]

Since $\mathbf{Q}/\mathbf{Z} = \varinjlim (1/n) \mathbf{Z}/\mathbf{Z}$ and $V = \varinjlim V[n]$ (as $V$ is torsion), it suffices to show that for $n > 0$ the natural map ${\rm{Hom}}(\mathbf{Q}/\mathbf{Z},V)/(n) \rightarrow V[n]$ defined by evaluation at $1/n \bmod \mathbf{Z}$ is an isomorphism (as then with a small diagram chase we can pass to the direct limit over more divisible $n$ to conclude).

But $V$ is an injective abelian group (as $V$ is divisible), so applying ${\rm{Hom}}(\cdot, V)$ to the exact sequence $$0 \rightarrow (1/n)\mathbf{Z}/\mathbf{Z} \rightarrow \mathbf{Q}/\mathbf{Z} \stackrel{n}{\rightarrow} \mathbf{Q}/\mathbf{Z} \rightarrow 0$$ yields exactly the desired isomorphism.

[In effect, this is a reformulation of the same argument as in Neil Strickland's answer, as I am masking some calculations implicit in the proof that divisible abeian groups are injective.]

Since $\mathbf{Q}/\mathbf{Z} = \varinjlim (1/n) \mathbf{Z}/\mathbf{Z}$ and $V = \varinjlim V[n]$ (as $V$ is torsion), it suffices to show that for $n > 0$ the natural map ${\rm{Hom}}(\mathbf{Q}/\mathbf{Z},V)/(n) \rightarrow V[n]$ defined by evaluation at $1/n \bmod \mathbf{Z}$ is an isomorphism (as then with a small diagram chase we can pass to the direct limit over more divisible $n$ to conclude).

But $V$ is an injective abelian group (as $V$ is divisible), so applying ${\rm{Hom}}(\cdot, V)$ to the exact sequence $$0 \rightarrow (1/n)\mathbf{Z}/\mathbf{Z} \rightarrow \mathbf{Q}/\mathbf{Z} \stackrel{n}{\rightarrow} \mathbf{Q}/\mathbf{Z} \rightarrow 0$$ yields exactly the desired isomorphism.

[In effect, this is a reformulation of the same argument as in Neil Strickland's answer, as I am masking some calculations implicit in the proof that divisible abeian groups are injective. I am also tacitly using that $\widehat{\mathbf{Z}}$ is the endomorphism ring of $\mathbf{Q}/\mathbf{Z}$, so tensoring over it against a torsion module is the same as tensoring over $\mathbf{Z}$.]