Here we hope to confirm the cohomology group of a manifold that behaves as some homogeneous spaces obtained from Lie groups. In particular with the coefficients of mod 2 or finite order 2 $\mathbb{Z}/2$. We focus on one example,
$$ H^2\left(\frac{O(10)}{U(5)}, \mathbb{Z}\right) $$ $$ H^2\left(\frac{O(10)}{U(5)}, \mathbb{Z}/2\right) $$
What I know so far is that $$ \pi_0\left(\frac{O(10)}{U(5)}\right)=\mathbb{Z}/2, \quad \pi_1\left(\frac{O(10)}{U(5)}\right)=0, \quad \pi_2\left(\frac{O(10)}{U(5)}\right)=\mathbb{Z}. $$
We also can use the universal coefficient theorem (UCT) such that $$ H^2(X,A)=Hom(H_2(X),A)\oplus Ext(H_1(X),A). $$
Am I correct to expect $$H^2\left(\frac{O(10)}{U(5)}, \mathbb{Z}\right)=\mathbb{Z}?$$ $$H^2\left(\frac{O(10)}{U(5)}, \mathbb{Z}/2\right)=\mathbb{Z}/2?$$ How do you demonstrate this statement?
One can answer the same question for $O(2n)/U(n)$ without any more effort.
The space $O(2n)/U(n)$ has two connected components, each of which is diffeomorphic to $SO(2n)/U(n)$, so $H^k(O(2n)/U(n); A) \cong H^k(SO(2n)/U(n); A)\oplus H^k(SO(2n)/U(n); A)$.
For $n = 1$, the space $SO(2)/U(1)$ is a point, so $H^2(O(2)/U(1); \mathbb{Z}) = 0$ and $H^2(O(2)/U(1); \mathbb{Z}_2) = 0$.
For $n > 1$, the space $SO(2n)/U(n)$ is simply connected with $\pi_2(SO(2n)/U(n)) \cong \mathbb{Z}$, so by the Hurewicz Theorem we have $H_1(SO(2n)/U(n); \mathbb{Z}) = 0$ and $H_2(SO(2n)/U(n); \mathbb{Z}) \cong \mathbb{Z}$. Therefore $H^2(SO(2n)/U(n); A) \cong \operatorname{Hom}(\mathbb{Z}, A)\oplus\operatorname{Ext}(0, A) \cong A$ by the Universal Coefficient Theorem, so
\begin{align*} H^2(O(2n)/U(n); \mathbb{Z}) &\cong \mathbb{Z}\oplus\mathbb{Z}\\ H^2(O(2n)/U(n); \mathbb{Z}_2) &\cong \mathbb{Z}_2\oplus\mathbb{Z}_2. \end{align*}