Equation of given curves, $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$
Let $\left(x_0, y_0\right)$ be point of intersection.
$\therefore$ Angle between curves $=$ Angle between tangent at point of intersection.
Now, $x^2-y^2=4$
Differentiate w.r.t. ' $x$ '
$$
2 x-2 y \frac{d y}{d x}=0 \quad \frac{d y}{d x}=\frac{x}{y}
$$

At point $\left(x_0, y_0\right)$
$$
m_1=\frac{x_0}{y_0} \text { and curve } x^2+y^2=4 \sqrt{2}
$$

Differentiate w.r.t. ' $x$ '
$$
2 x+2 y \frac{d y}{d x}=0 \quad \frac{d y}{d x}=-\frac{x}{y}
$$

At point $\left(x_0, y_0\right)$
$$
m_2=-\frac{x_0}{y_0}
$$


Angle between tangents $\theta=\tan ^{-1}\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$
$$
\begin{aligned}
& \theta=\tan ^{-1}\left|\frac{\frac{x_0}{y_0}+\frac{x_0}{y_0}}{1+\frac{x_0}{y_0}\left(-\frac{x_0}{y_0}\right)}\right| \Rightarrow \theta=\tan ^{-1}\left|\frac{2 \frac{x_0}{y_0}}{\frac{y_0^2-x_0^2}{y_0^2}}\right| \\
& \Rightarrow \quad \theta=\tan ^{-1}\left(\frac{2 x_0 y_0}{y_0^2-x_0^2}\right)=\tan ^{-1}\left(\frac{2 x_0 y_0}{4}\right)
\end{aligned}
$$
[given, $y^2-x^2=4$ satisfied point $\left(x_0, y_0\right)$ ]
$$
=\tan ^{-1}\left(\frac{x_0 y_0}{2}\right)
$$


Point $\left(x_0, y_0\right)$ lie on curves, so
$$
\begin{array}{rlrl}
& & x_0^2-y_0^2 & =4 \text { and } x_0^2+y_0^2=4 \sqrt{2} \\
\Rightarrow & & x_0^2 & =2(1+\sqrt{2}) \text { and } y_0^2=2(\sqrt{2}-1) \\
& \text { So, } \quad & x_0^2 y_0^2 & =4(\sqrt{2}+1)(\sqrt{2}-1)=4(2-1)=4 \\
x_0 y_0 & =2
\end{array}
$$

Put in Eq. (i),
$$
\theta=\tan ^{-1}\left(\frac{2}{2}\right) \quad \theta=\tan ^{-1}(1) \Rightarrow \theta=\frac{\pi}{4}
$$