I think that the best way to compute this cohomology is the following. The homogeneous spaces you are looking at are all of the form $X=G/M$ where $G$ is compact and $M$ is a connected subgroup of the same rank, so they have a common maximal torus $T$. Let us first look at the equivariant cohomology $H^*_G(X)$ (say, with $\mathbb C$-coefficients). It is obvious that it is the same as $H^*_M(pt)$ (here $pt$ denote "the point") which is known to be $Sym(\mathfrak t^*)^{W_M}$; here $Sym$ means "symmetric algebra", $\mathfrak t$ is the complexified Lie algebra of $T$ and $W_M$ means the Weyl group of $M$. By abstract nonsense it is clear that $H^*(X)$

is just $H^*_G(X)\underset{H^*_G(pt)}\otimes {\mathbb C}$, where $\otimes$ in principle means "derived tensor product". However, you can show that in the above cases $H^*_G(X)$ is always free over $H^*_G(pt)=Sym(\mathfrak t^*)^{W_G}$, hence you finally get

that $H^*(X)$

is equal to $Sym(\mathfrak t^*)^{W_M}\underset{Sym(\mathfrak t^*)^{W_G}}\otimes{\mathbb C}$.

Note that $W_M$ is a subgroup of $W_G$.

I think that the best way to compute this cohomology is the following. The homogeneous spaces you are looking at are all of the form $X=G/M$ where $G$ is compact and $M$ is a connected subgroup of $G$ Let $T_M$ be a maximal torus of $M$, embedded into a maximal torus $T_G$ of $G$ and let $\mathfrak t_M$ $\mathfrak t_G$ be the corresponding (complexified) Lie algebras. Let us first look at the equivariant cohomology $H^*_G(X)$ (say, with $\mathbb C$-coefficients). It is obvious that it is the same as $H^*_M(pt)$ (here $pt$ denote "the point") which is known to be $Sym(\mathfrak t_M^*)^{W_M}$; here $Sym$ means "symmetric algebra", and $W_M$ means the Weyl group of $M$. By abstract nonsense it is clear that $H^*(X)$

is just $H^*_G(X)\underset{H^*_G(pt)}\otimes {\mathbb C}$, where $\otimes$ in principle means "derived tensor product". If $M$ and $G$ have the same rank, then you can show that $H^*_G(X)$ is always free over $H^*_G(pt)=Sym(\mathfrak t^*)^{W_G}$ (here I denote $\mathfrak t=\mathfrak t_M=\mathfrak t_G$) hence you finally get

that $H^*(X)$

is equal to $Sym(\mathfrak t^*)^{W_M}\underset{Sym(\mathfrak t^*)^{W_G}}\otimes{\mathbb C}$.

In the general case, you need to compute the above derived tensor product, which in every specific case is usually easy to do.

I think that the best way to compute this cohomology is the following. The homogeneous spaces you are looking at are all of the form $X=G/M$ where $G$ is compact and $M$ is a connected subgroup of the same rank, so they have a common maximal torus $T$. Let us first look at the equivariant cohomology $H^*_G(X)$ (say, with $\mathbb C$-coefficients). It is obvious that it is the same as $H^*_M(pt)$ (here $pt$ denote "the point") which is known to be $Sym(\mathfrak t^*)^{W_M}$; here $Sym$ means "symmetric algebra", $\mathfrak t$ is the complexified Lie algebra of $T$ and $W_M$ means the Weyl group of $M$. By abstract nonsense it is clear that $H^*(X)$

is just $H^*_G(X)\underset{H^*_G(pt)}\otimes {\mathbb C}$, where $\otimes$ in principle means "derived tensor product". However, you can show that in the above cases $H^*_G(X)$ is always free over $H^*_G(pt)=Sym(\mathfrak t^*)^{W_G}$, hence you finally get

that $H^*(X)$

is equal to $Sym(\mathfrak t^*)^{W_M}\underset{Sym(\mathfrak t^*)^{W_G}}\otimes{\mathbb C}$.

Note that $W_G$ is a subgroup of $W_M$.

I think that the best way to compute this cohomology is the following. The homogeneous spaces you are looking at are all of the form $X=G/M$ where $G$ is compact and $M$ is a connected subgroup of the same rank, so they have a common maximal torus $T$. Let us first look at the equivariant cohomology $H^*_G(X)$ (say, with $\mathbb C$-coefficients). It is obvious that it is the same as $H^*_M(pt)$ (here $pt$ denote "the point") which is known to be $Sym(\mathfrak t^*)^{W_M}$; here $Sym$ means "symmetric algebra", $\mathfrak t$ is the complexified Lie algebra of $T$ and $W_M$ means the Weyl group of $M$. By abstract nonsense it is clear that $H^*(X)$

is just $H^*_G(X)\underset{H^*_G(pt)}\otimes {\mathbb C}$, where $\otimes$ in principle means "derived tensor product". However, you can show that in the above cases $H^*_G(X)$ is always free over $H^*_G(pt)=Sym(\mathfrak t^*)^{W_G}$, hence you finally get

that $H^*(X)$

is equal to $Sym(\mathfrak t^*)^{W_M}\underset{Sym(\mathfrak t^*)^{W_G}}\otimes{\mathbb C}$.

Note that $W_M$ is a subgroup of $W_G$.