Added (by Ralph): Here are some more details for the spectral sequences. We know $K(\mathbb{Z}/2,1)=\mathbb{R}P^\infty$ and $K(\mathbb{Z},2)=\mathbb{C}P^\infty$ and \mathbb{Z} & p=0 \newline \mathbb{Z}_2 & p > 0 \text{ even }\;\;, \newline 0 & p > 0 \text{ odd }M & p> 0 \text{ even} \newline 0 & p > 0 \text{ odd } where $\mathbb{Z}_2 := \mathbb{Z}/2$ and $M$ are trivial coefficients. Since $\mathbb{Z}_2$ has only two elements, the coefficient system in the LS spectral sequence of the first fibration is trivial and we obtain for $E_2^{p,q}(1)=H^P(\mathbb{C}P^\infty;H^q(\mathbb{R}P^\infty;\mathbb{Z}))$: \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \newline
\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & \cdots \newline0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & \cdots\newline0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \cdots\newlineNow, for positional reasons, $E_2(1)=E_\infty(1)$.
As the 2nd fibration is trivial, the coefficient system in its LS spectral sequence is also trivial. Hence both spectral sequences agree (in all terms), while the cohomologies differ: $H^p(\mathbb{C}P^\infty;\mathbb{Z})$ is described above and $$H^p(\mathbb{C}P^\infty \times \mathbb{R}P^\infty;\mathbb{Z})=\begin{cases}\mathbb{Z} \oplus \mathbb{Z}_2^n & p= 2n \newline 0 & p \text{ odd.}
\end{cases}$$
