As the following product is a bit unfamiliar to me: 



> 
> How do we compute the rational cohomology ring $H^\*(M;\mathbb{Q})$ of the product of Lie groups: 
>  $M=SO(n\_1)\times U(n\_2)\times SU(n\_3)\times (Sp(n\_4)\cdot Sp(1))\times Sp(n\_5)\times G\_2\times Spin(7)\times \left(\frac{SU(n\_6+1)}{S(U(1)\times U(n\_6))}\right)\times \left(\frac{SO(n\_7+1)}{SO(1)\times SO(n\_7)}\right)\times \left(\frac{Sp(n\_8+1)}{Sp(1)\times Sp(n\_8)}\right)\times \left(\frac{F\_4}{Spin(9)}\right)?$
> 
> 
> 


*Thoughts*: Of course, $H^\*(M;\mathbb{Q})$ is: $$H^\*(M;\mathbb{Q})\cong H^\*(SO(n\_1);\mathbb{Q})\otimes H^\*(U(n\_2);\mathbb{Q})\otimes H^\*(SU(n\_3);\mathbb{Q})\otimes...\otimes H^\*(F^4/Spin(9);\mathbb{Q})$$ by the Kunneth formula. This could be possibly simplified by first computing the rationalization $M\_\mathbb{Q}$, which would help us in finding the equivariant cohomology ring $H^\*\_{\mathbb{Q}}(M;\mathbb{Q})=H^\*(E\mathbb{Q}\times\_{\mathbb{Q}}M;\mathbb{Q})$, noting that the rationalization $M\_\mathbb{Q}$ is a product of Eilenberg–MacLane spaces (whose rational cohomology rings are better known-the problem is in finding such a product decomposition).



Compute the rational cohomology ring $H^*(M;\mathbb{Q})$ of a product of Lie groups (and their quotients)
See Section 3 in Fok's [nice survey paper.](http://pi.math.cornell.edu/~ckfok/Cohomology_Lie_groups.pdf) (Cohomology and K-Theory of compact Lie groups), and all will be revealed.


NOTE: I just noticed that (as pointed out by Arun Debray) that yours is not a product of Lie groups, so maybe not all will be revealed.


