Answer
$$\theta=\frac{\pi}{4}.$$
Work Step by Step
\begin{align*}
\cos \theta &=\frac{\langle 2,\sqrt{2}\rangle \cdot \langle 1+\sqrt{2},1-\sqrt{2}\rangle}{\|\langle 2,\sqrt{2}\rangle\|\|\langle 1+\sqrt{2},1-\sqrt{2}\rangle\|}\\
&=\frac{2+2\sqrt{2}+\sqrt{2}-2}{\sqrt{4+2}\sqrt{1+2\sqrt{2}+2+1-2\sqrt{2}+2}}\\
&=\frac{3\sqrt{2}}{\sqrt{6}\sqrt{6}}\\
&=\frac{\sqrt{2}}{2}
\end{align*}
That is $$\theta=\cos^{-1}\frac{\sqrt{2}}{2}=\frac{\pi}{4}.$$
