$H^*(G_{2m+1}(\mathbb{R}^\infty);\mathbb{Z}/ 2)$ forms a cochain complex with respect to the differential operator $\mathrm{Sq}^1$. Compute the cohomology.
By the Bockstein exact sequence $$\dots \longrightarrow H^j(\; \; \; ; \mathbb{Z}) \overset{2}{\longrightarrow} H^j(\; \; \; ; \mathbb{Z}) \overset{\rho}{\longrightarrow} H^j(\; \; \; ; \mathbb{Z}/2) \overset{\beta}{\longrightarrow}H^{j+1}(\; \; \; ; \mathbb{Z}) \overset{2}{\longrightarrow}\dots ,$$ prove that $H^*(G_{2m+1}(\mathbb{R}^\infty);\mathbb{Z})$ splits additively as the direct sum of the polynomial ring $\mathbb{Z}[p_1,\dots,p_m]$ where $p_i$ is the Pontrjagin class. Notice that $\rho \circ \beta = \mathrm{Sq}^1$
Prove the analogous statements for $G_{2m}(\mathbb{R}^\infty)$ and $\tilde{G}_{n}(R^\infty)$
To compute the cohomology, it's convient to set $$H^*(G_{2m+1}(\mathbb{R}^\infty);\mathbb{Z}/ 2) = \mathbb{Z}/2[w_1,w_2,\mathrm{Sq}^1 (w_2),\dots,w_{2m},\mathrm{Sq}^1(w_{2m})]=\mathbb{Z}/2[w_1] \otimes {\mathbb{Z}/2}[w_2,\mathrm{Sq}^1 (w_2),\dots,w_{2m}, \mathrm{Sq}^1(w_{2m})].$$ We can see by induction that if the linear combination of monomials in ${\mathbb{Z}/2}[w_2,\mathrm{Sq}^1 (w_2),\dots,w_{2m}, \mathrm{Sq}^1(w_{2m})]$ involving some $\mathrm{Sq}^1(w_{2i})$ in each term is annihilated by $\mathrm{Sq}^1$ then it is in the image of $\mathrm{Sq}^1$. And obviously cohomology ring of $\mathbb{Z}/2[w_1]$ is $\mathbb{Z}/2$. THerefore the total cohomology ring is $\mathbb{Z}/2[w_{wi}^2,i=1,\dots,m]$. Similarly for $G_{2m}$. The oriented case is more discrete and may be more easy.
Since $\rho \circ \beta = \mathrm{Sq}^1$ we can see that $$\rho(\mathrm{Im}(\beta)) = \mathrm{Im}(\mathrm{Sq}^1) \subset \mathrm{Ker}(\beta) \subset \mathrm{Ker}(\mathrm{Sq}^1) = \mathrm{Im}(\mathrm{Sq}^1) + \mathbb{Z}/2[w_{wi}^2,i=1,\dots,m] = \rho(\mathbb{Z}[p_1,\dots,p_m]\oplus \mathrm{Im}(\beta)).$$ After that it is hard to go on.
Any hit could help.