As it has been a good many years since I've done any vector manipulation, some help to jolt my memory would be good.
The two plane waves are
$\vec{E_{1}} = \hat{e_{1}}|\vec{E}_{1}| e^{i(\vec{k}\vec{r} - \omega t)}e^{i\phi_{1}}$
and
$\vec{E_{2}} = \hat{e_{2}}|\vec{E}_{2}| e^{i(\vec{k}\vec{r} - \omega t)}e^{i\phi_{2}}$
where
$\vec{k} = \langle k_{x}, k_{y}, k_{z} \rangle$, $\vec{r} = \langle r_{x}, r_{y}, r_{z} \rangle$.
I want to show that $|\vec{E}|^{2} = |E_{1}|^{2} + |E_{2}|^{2} + 2 |\vec{E}_{1}| |\vec{E}_{2}| cos(\phi)$
The square of the magnitude for $\vec{E}_{1}$ gives $|\vec{E}_{1}|^{2} = |\vec{E}_{1}| |e^{i(\vec{k}\vec{r} - \omega t)}e^{i\phi_{1}}| = |\vec{E}_{1}| |e^{-(\vec{k}\vec{r} - \omega t)\phi_{1}}|$
and the square of the magnitude for $\vec{E}_{2}$ gives $|\vec{E}_{2}|^{2} = |\vec{E}_{2}| |e^{i(\vec{k}\vec{r} - \omega t)}e^{i\phi_{2}}| = |\vec{E}_{2}| |e^{-(\vec{k}\vec{r} - \omega t)\phi_{2}}|$
Any further help or hints are appreciated.
This is not a complete answer because I don't know how to account for the angle, but hopefully it still helps. For any vector sum $E=E_1+E_2$, we can write
\begin{align} |E|^2&=\langle E_1+E_2, E_1+E_2 \rangle=\langle E_1, E_1+E_2 \rangle +\langle E_2, E_1+E_2 \rangle\\ &=\langle E_1, E_1 \rangle+\langle E_1, E_2 \rangle +\langle E_2, E_1 \rangle +\langle E_2, E_2 \rangle\\ &=|E_1|^2 + |E_2|^2+2\langle E_1, E_2 \rangle\\ &=|E_1|^2 + |E_2|^2+2|E_1| |E_2| \cos(\phi), \end{align}
where $\phi$ is the angle between $E_1$ and $E_2$. But something is fishy in the problem statement, because if $\hat{e_1}$ and $\hat{e_2}$ are the standard basis vectors in $\mathbb{R}^2$, then $\cos(\phi)=\cos(\pi/2)=0$.