Let
$$
\begin{aligned}
f(x)= & \cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right) \\
= & \left(\cos x \cdot \cos \frac{\pi}{3}-\sin x \cdot \sin \frac{\pi}{3}\right) \\
& +2 \sqrt{2}\left(\sin x \cos \frac{\pi}{3}+\cos x \sin \frac{\pi}{3}\right)
\end{aligned}
$$

$$
\begin{aligned}
&=\left(\cos x \cdot \frac{1}{2}-\sin x \frac{\sqrt{3}}{2}\right) \\
&+2 \sqrt{2}\left(\sin x \cdot \frac{1}{2}+\cos x \cdot \frac{\sqrt{3}}{2}\right) \\
&= \frac{\cos x}{2}-\frac{\sqrt{3} \sin x}{2}+\sqrt{2} \sin x+\sqrt{6} \cos x \\
& f(x)= \cos x\left(\frac{1}{2}+\sqrt{6}\right)+\sin x\left(\sqrt{2}-\frac{\sqrt{3}}{2}\right) \\
& \therefore \quad a= \frac{1}{2}+\sqrt{6} \text { and } b=\sqrt{2}-\frac{\sqrt{3}}{2} \\
& \sqrt{a^2+b^2}= \sqrt{\left(\frac{1}{2}+\sqrt{6}\right)^2+\left(\sqrt{2}-\frac{\sqrt{3}}{2}\right)^2}=\sqrt{9} \\
& \sqrt{a^2+b^2}= 3
\end{aligned}
$$
$\therefore$ Maximum value of $f(x)=\sqrt{a^2+b^2}=3$
Minimum value of $f(x)=-\sqrt{a^2+b^2}=-3$
Hence, option (3) is correct.