Short answer: $H^i(\mathbb{Q}/\mathbb{Z}) = \begin{cases} \mathbb{Z}, i = 0,\\ 0, i \equiv 1 \mod 2,\\ A, i\equiv 0 \mod 2, i>1. \end{cases}$

Where $A$ fits into a short exact sequence $\mathbb{Z}\to A\to H^2(\mathbb{Q})$.

Explanation:

We can use the short exact sequence to compute the cohomology of $\mathbb{Q}$ from the cohomology of $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{Z}$.

If we look at the spectral sequence we see that only the first two rows are non-zero since the $(p, q)$ entry is $H^p(\mathbb{Q}/\mathbb{Z}; H^q(\mathbb{Z}; \mathbb{Z}))$ which is non-zero only if $H^q(\mathbb{Z})\neq 0$ only if $q = 0, 1$.

Thus we end up with a long exact sequence of cohomology groups $$0 \to H^1(\mathbb{Q}/\mathbb{Z})\to H^1(\mathbb{Q})\to H^0(\mathbb{Q}/\mathbb{Z})\to \dots \to H^i(\mathbb{Q})\to H^{i-1}(\mathbb{Q}/\mathbb{Z}) \to H^{i+1}(\mathbb{Q}/\mathbb{Z})\to H^{i+1}(\mathbb{Q})\to\dots.$$

Now $\mathbb{Q}$ has cohomological dimension 2. So we obtain $H^i(\mathbb{Q}/\mathbb{Z}) = H^{i+2}(\mathbb{Q}/\mathbb{Z})$ for $i>2$.

For the low dimensional cases since $\mathbb{Q}/\mathbb{Z}$ is torsion we see $H^1(\mathbb{Q}/\mathbb{Z})$ vanishes.

Also $H^1(\mathbb{Q})$ vanishes so we get a short exact sequence $0\to H^0(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q})\to 0$. Hence the even terms.

We also see that a portion of the long exact sequence is: $$\dots\to H^{1}(\mathbb{Q}/\mathbb{Z}) \to H^{3}(\mathbb{Q}/\mathbb{Z})\to H^{3}(\mathbb{Q})\to \dots.$$ So $H^3(\mathbb{Q}/\mathbb{Z})$ vanishes.

Unfortunately, I do not understand $H^2(\mathbb{Q})$, let alone an extension of it by $\mathbb{Z}$. I believe the universal coefficient theorem shows that $H^2(\mathbb{Q}\cong $ Ext$(\mathbb{Q}, \mathbb{Z})$ which is described in some detail on page 5 of these notes: http://math.jhu.edu/~jmb/note/torext.pdf

Short answer: $$ H^i(\mathbb{Q}/\mathbb{Z}) = \begin{cases} \mathbb{Z}, & i = 0,\\ 0, & i \equiv 1 \mod 2,\\ A, & i\equiv 0 \mod 2, i>1, \end{cases}$$ where $A$ fits into a short exact sequence $\mathbb{Z}\to A\to H^2(\mathbb{Q})$.

Explanation:

We can use the short exact sequence to compute the cohomology of $\mathbb{Q}$ from the cohomology of $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{Z}$.

If we look at the spectral sequence we see that only the first two rows are non-zero since the $(p, q)$ entry is $H^p(\mathbb{Q}/\mathbb{Z}; H^q(\mathbb{Z}; \mathbb{Z}))$ which is non-zero only if $H^q(\mathbb{Z})\neq 0$ only if $q = 0, 1$.

Thus we end up with a long exact sequence of cohomology groups $$0 \to H^1(\mathbb{Q}/\mathbb{Z})\to H^1(\mathbb{Q})\to H^0(\mathbb{Q}/\mathbb{Z})\to \dotsb \to H^i(\mathbb{Q})\to H^{i-1}(\mathbb{Q}/\mathbb{Z}) \to H^{i+1}(\mathbb{Q}/\mathbb{Z})\to H^{i+1}(\mathbb{Q})\to\dotsb.$$

Now $\mathbb{Q}$ has cohomological dimension 2. So we obtain $H^i(\mathbb{Q}/\mathbb{Z}) = H^{i+2}(\mathbb{Q}/\mathbb{Z})$ for $i>2$.

For the low dimensional cases since $\mathbb{Q}/\mathbb{Z}$ is torsion we see $H^1(\mathbb{Q}/\mathbb{Z})$ vanishes.

Also $H^1(\mathbb{Q})$ vanishes so we get a short exact sequence $0\to H^0(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q})\to 0$. Hence the even terms.

We also see that a portion of the long exact sequence is: $$\dotsb\to H^{1}(\mathbb{Q}/\mathbb{Z}) \to H^{3}(\mathbb{Q}/\mathbb{Z})\to H^{3}(\mathbb{Q})\to \dotsb.$$ So $H^3(\mathbb{Q}/\mathbb{Z})$ vanishes.

Unfortunately, I do not understand $H^2(\mathbb{Q})$, let alone an extension of it by $\mathbb{Z}$. I believe the universal coefficient theorem shows that $H^2(\mathbb{Q})\cong \operatorname{Ext}(\mathbb{Q}, \mathbb{Z})$, which is described in some detail on page 5 of the notes Boardman - Some common Tor and Ext groups.

Short answer: $H^i(\mathbb{Q}/\mathbb{Z}) = \begin{cases} \mathbb{Z}, i = 0,\\ 0, i \equiv 1 \mod 2,\\ A, i\equiv 0 \mod 2, i>1. \end{cases}$

Where $A$ fits into a short exact sequence $0\to \mathbb{Z}\to A\to H^2(\mathbb{Q})\to 0$.

Explanation:

We can use the short exact sequence to compute the cohomology of $\mathbb{Q}$ from the cohomology of $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{Z}$.

First note that $\mathbb{Q}$ has cohomological dimension 2. We can see that it has homological dimension 1 since it is a direct limit of cyclic groups and homology commutes with direct limits. Then Bieri [1] shows that the cohomological dimension is $\leq 2$. Groups of cohomological dimension 1 are free so it must be that $\mathbb{Q}$ has cohomological dimension 2. Thus, $H^i(\mathbb{Q}; \mathbb{Z}) = 0$ for $i\geq 3$.

Now if we look at the spectral sequence we see that only the first two rows are non-zero since the $(p, q)$ entry is $H^p(\mathbb{Q}/\mathbb{Z}; H^q(\mathbb{Z}; \mathbb{Z}))$ which is non-zero only if $H^q(\mathbb{Z})\neq 0$ only if $q = 0, 1$.

Since the extension is central the action of $\mathbb{Q}/\mathbb{Z}$ on $H^q(\mathbb{Z})$ is trivial and so the $(p, q)$ term of the spectral sequence is $H^p(\mathbb{Q}/\mathbb{Z}; \mathbb{Z})$ for $q= 0, 1$ and trivial otherwise.

For what follows I will be taking all cohomology groups with trivial $\mathbb{Z}$ coefficients and supress them throughout.

Since there are only non-trivial terms on the first two rows of the spectral sequence we see that it stabilises after the third page. On the second page we have maps $H^i(\mathbb{Q}/\mathbb{Z})\to H^{i+2}(\mathbb{Q}/\mathbb{Z})$.

Since $H^i(\mathbb{Q})$ vanishes for $i\geq 3$ we know that the $(p, q)$-term of the third page vanishes for $p+q\geq 3$. Thus for $i\geq 2$, the map $H^i(\mathbb{Q}/\mathbb{Z})\to H^{i+2}(\mathbb{Q}/\mathbb{Z})$ from the second page is an isomorphism.

Thus $H^i(\mathbb{Q}/\mathbb{Z})\cong H^{i+2}(\mathbb{Q}/\mathbb{Z})$ for $i\geq 2$.

Since $\mathbb{Q}/\mathbb{Z}$ is torsion we have that $H^1(\mathbb{Q}/\mathbb{Z}) = 0$.

Also since the term in the $(3, 0)$ position on the third page of the spectral sequence is trivial we see that the map $H^1(\mathbb{Q}/\mathbb{Z})\to H^{3}(\mathbb{Q}/\mathbb{Z})$ is surjective and so $H^3(\mathbb{Q}/\mathbb{Z}) = 0$.

Now we are left to compute $H^2(\mathbb{Q}/\mathbb{Z})$. From the second page of the spectral sequence we have the map $\mathbb{Z} = H^0(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q}/\mathbb{Z})$. Since $H^1(\mathbb{Q})$ vanishes this map is injective. Moreover, the cokernel of this map is $H^2(\mathbb{Q})$. Thus $H^2(\mathbb{Q}/\mathbb{Z})/\mathbb{Z} \cong H^2(\mathbb{Q})$.

Now I apologise, for this is where my answer stops. Unfortunately, I do not understand $H^2(\mathbb{Q})$, let alone an extension of it by $\mathbb{Z}$. Using the universal coefficient theorem we see $H^2(\mathbb{Q})\cong $ Ext$(\mathbb{Q}, \mathbb{Z})$ which is described in some detail on page 5 of these notes: http://math.jhu.edu/~jmb/note/torext.pdf

[1] R. Bieri, Normal subgroups in duality groups and in groups of cohomological dimension 2. J. Pure Appl. Algebra 7 (1976), no. 1, 35–51.

Short answer: $H^i(\mathbb{Q}/\mathbb{Z}) = \begin{cases} \mathbb{Z}, i = 0,\\ 0, i \equiv 1 \mod 2,\\ A, i\equiv 0 \mod 2, i>1. \end{cases}$

Where $A$ fits into a short exact sequence $\mathbb{Z}\to A\to H^2(\mathbb{Q})$.

Explanation:

We can use the short exact sequence to compute the cohomology of $\mathbb{Q}$ from the cohomology of $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{Z}$.

If we look at the spectral sequence we see that only the first two rows are non-zero since the $(p, q)$ entry is $H^p(\mathbb{Q}/\mathbb{Z}; H^q(\mathbb{Z}; \mathbb{Z}))$ which is non-zero only if $H^q(\mathbb{Z})\neq 0$ only if $q = 0, 1$.

Thus we end up with a long exact sequence of cohomology groups $$0 \to H^1(\mathbb{Q}/\mathbb{Z})\to H^1(\mathbb{Q})\to H^0(\mathbb{Q}/\mathbb{Z})\to \dots \to H^i(\mathbb{Q})\to H^{i-1}(\mathbb{Q}/\mathbb{Z}) \to H^{i+1}(\mathbb{Q}/\mathbb{Z})\to H^{i+1}(\mathbb{Q})\to\dots.$$

Now $\mathbb{Q}$ has cohomological dimension 2. So we obtain $H^i(\mathbb{Q}/\mathbb{Z}) = H^{i+2}(\mathbb{Q}/\mathbb{Z})$ for $i>2$.

For the low dimensional cases since $\mathbb{Q}/\mathbb{Z}$ is torsion we see $H^1(\mathbb{Q}/\mathbb{Z})$ vanishes.

Also $H^1(\mathbb{Q})$ vanishes so we get a short exact sequence $0\to H^0(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q})\to 0$. Hence the even terms.

We also see that a portion of the long exact sequence is: $$\dots\to H^{1}(\mathbb{Q}/\mathbb{Z}) \to H^{3}(\mathbb{Q}/\mathbb{Z})\to H^{3}(\mathbb{Q})\to \dots.$$ So $H^3(\mathbb{Q}/\mathbb{Z})$ vanishes.

Unfortunately, I do not understand $H^2(\mathbb{Q})$, let alone an extension of it by $\mathbb{Z}$. I believe the universal coefficient theorem shows that $H^2(\mathbb{Q}\cong $ Ext$(\mathbb{Q}, \mathbb{Z})$ which is described in some detail on page 5 of these notes: http://math.jhu.edu/~jmb/note/torext.pdf