= Solution
Put $D=t_1+t_2$, $a=t_1\rho_{\rm FM}^L+t_2\rho_{\rm NM}$ and $b=t_1\rho_{\rm FM}^H+t_2\rho_{\rm NM}$. In the <CIP thickness-averaged two-current model>, assume spin is conserved and electron trajectories sample the layers, so each channel's effective resistivity is the thickness average: $\bar\rho_L=a/D$, $\bar\rho_H=b/D$. This uses averaged relaxation rates with comparable transport weights, neglecting additional interface resistance. Here $\rho_{\rm NM}$ is the per-spin-channel resistivity, twice the measured nonmagnetic bulk resistivity. The stated long <mean free path> motivates sampling more than one layer.
A four-layer period has thickness $2D$. For an in-plane square, $R_\sigma=\bar\rho_\sigma L/(2DL)=\bar\rho_\sigma/(2D)$, hence the two parallel spin conductances add:
$$
\boxed{G_P=\frac1{R_{\uparrow,\uparrow}}=2D^2\left(\frac1a+\frac1b\right).}
$$
The square's side $L$ cancels. The layers are not being treated as four macroscopic resistors in series along the in-plane current; the averaging is over the traversing carriers' scattering histories.
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