Explicitly, as a free module over the ground ring $\mathbb{M}_2$:
$$H^{\ast,\ast}(\text{Gr}_4(\mathbb{R}^{8,1}))=\mathbb{M}_2\oplus\Sigma^{1,1}\mathbb{M}_2\oplus\Sigma^{2,1}\mathbb{M}_2\oplus\Sigma^{3,1}\mathbb{M}_2\oplus\Sigma^{4,1}\mathbb{M}_2\oplus\Sigma^{4,2}\mathbb{M}_2\oplus\Sigma^{5,1}\mathbb{M}_2\oplus\Sigma^{5,2}\mathbb{M}_2\oplus\Sigma^{6,1}\mathbb{M}_2\oplus\Sigma^{6,2}\mathbb{M}_2\oplus\Sigma^{7,1}\mathbb{M}_2\oplus\Sigma^{7,2}\mathbb{M}_2\oplus\Sigma^{8,2}\mathbb{M}_2\oplus\Sigma^{9,2}\mathbb{M}_2\oplus\Sigma^{9,3}\mathbb{M}_2\oplus\Sigma^{10,2}\mathbb{M}_2\oplus\Sigma^{10,3}\mathbb{M}_2\oplus\Sigma^{11,2}\mathbb{M}_2\oplus\Sigma^{11,3}\mathbb{M}_2\oplus\Sigma^{12,2}\mathbb{M}_2\oplus\Sigma^{12,3}\mathbb{M}_2\oplus\Sigma^{13,3}\mathbb{M}_2\oplus\Sigma^{14,3}\mathbb{M}_2\oplus\Sigma^{15,3}\mathbb{M}_2\oplus\Sigma^{16,4}\mathbb{M}_2.$$
The AutoKron can't determine the cohomology of $\text{Gr}_4(\mathbb{R}^{8,2})$.