= Solution
Let $u_n$ be the rationalized universal class of degree $n$. \b[The <rational cohomology of an integral Eilenberg–MacLane space> is]
$$
\boxed{H^*(K(\mathbb Z,n);\mathbb Q)\cong
\begin{cases}
\Lambda_{\mathbb Q}(u_n),&n\text{ odd},\\
\mathbb Q[u_n],&n\text{ even}.
\end{cases}}
$$
Here $\Lambda$ is an <exterior algebra>. For $n=1$, $K(\mathbb Z,1)\simeq S^1$. The standard path-loop spectral-sequence calculation supplies the induction: in
$$
K(\mathbb Z,n-1)\longrightarrow PK(\mathbb Z,n)
\longrightarrow K(\mathbb Z,n),
$$
the total space is contractible and the fundamental fiber class transgresses to $u_n$. An odd exterior fiber generator gives an even polynomial base generator. An even polynomial fiber generator gives an odd exterior base generator; the differential on its $k$th power has coefficient $k$, which is invertible over $\mathbb Q$. The multiplicative spectral sequence then has no remaining positive-degree classes in the total space. This is the rational transgression calculation for <Eilenberg–MacLane spaces>; it includes the absence of additional base generators.
For $m=2n+1\geq3$, choose $S^m\to K(\mathbb Z,m)$ representing the integral fundamental class. The ring calculation shows that this map is a rational <homology> equivalence: both spaces have rational <cohomology> only in degrees zero and $m$. The <rational Whitehead theorem> for <simply connected> spaces identifies their rational <homotopy groups>. Since the target has only $\pi_m=\mathbb Z$, \b[the <rational homotopy groups of a sphere> in odd dimension are]
$$
\boxed{\pi_i(S^{2n+1})\otimes\mathbb Q=
\begin{cases}
\mathbb Q,&i=2n+1,\\
0,&i\ne2n+1.
\end{cases}}
$$
For $n=0$, this follows directly from $\pi_1(S^1)=\mathbb Z$ and the contractible <universal cover> of the circle, which makes every higher <homotopy group> zero.
Let $M=\mathbb{CP}^2\#\mathbb{CP}^2$, with the standard complex orientations. It is <simply connected> by the <Seifert-van Kampen theorem> applied to the punctured summands. Classes $a,b\in H^2(M;\mathbb Q)$ can be chosen from the two summands. Their cross product vanishes, while their squares equal the oriented top class:
$$
ab=0,\qquad a^2=b^2=\omega.
$$
Thus the <cohomology ring of the connected sum of two complex projective planes> is
$$
H^*(M;\mathbb Q)=\mathbb Q[a,b]/(ab,a^2-b^2),
\qquad |a|=|b|=2.
$$
The two relations also kill all cubic monomials, so its dimensions are $1,2,1$ in degrees $0,2,4$ and zero otherwise.
We use the <Sullivan minimal model> dictionary: for a <simply connected> finite-type space, the dual of its degree-$i$ generator space is $\pi_i(M)\otimes\mathbb Q$. The following free graded-commutative differential algebra is the <Sullivan model of the connected sum of two complex projective planes>:
$$
\boxed{\mathcal M=(\Lambda(a_2,b_2,r_3,s_3),d),
\quad da=db=0,\quad dr=ab,\quad ds=a^2-b^2.}
$$
It is minimal because all differentials of generators are decomposable.
To verify that no further generators are required, observe that $ab,a^2-b^2$ is a <regular sequence> in $\mathbb Q[a,b]$. The first polynomial is a nonzerodivisor. If $(a^2-b^2)h$ is divisible by $ab$, restricting to each coordinate axis forces $h$ to vanish on both axes, hence to be divisible by $ab$. The second polynomial is therefore a nonzerodivisor modulo the first. The <Koszul complex> of this <regular sequence> is exactly $\mathcal M$, so its <cohomology> is the quotient ring above, with no additional odd <cohomology>.
For completeness, choose rational polynomial forms representing $a,b$ on $M$. Their product and the difference of their squares are exact; choose degree-three primitives for them. Sending $r,s$ to those primitives defines a differential-algebra map from $\mathcal M$ to the rational polynomial forms on $M$. It induces the specified cohomology-ring isomorphism and hence is a <quasi-isomorphism>. This verifies the model directly, rather than assuming that a <cohomology> presentation alone automatically determines all rational <homotopy>.
The model has exactly two degree-two and two degree-three generators. Consequently \b[the <rational homotopy groups of the connected sum of two complex projective planes> are]
$$
\boxed{\pi_i(\mathbb{CP}^2\#\mathbb{CP}^2)\otimes\mathbb Q=
\begin{cases}
\mathbb Q^2,&i=2,3,\\
0,&i=1\text{ or }i\geq4.
\end{cases}}
$$
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