We have, $\overrightarrow{a}=\left(\overrightarrow{b}·\overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{b}·\overrightarrow{b}\right)\overrightarrow{c}$

$=1.2\cos \theta \overrightarrow{b}-\overrightarrow{c}$

$\Rightarrow \overrightarrow{a}=2\cos \theta \overrightarrow{b}-\overrightarrow{c}$

${\left|\overrightarrow{a}\right|}^2={\left|2\cos \theta \overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2-2.2\cos \theta \left(\overrightarrow{b}·\overrightarrow{c}\right)$

${\left|\overrightarrow{a}\right|}^2={\left(2\cos \theta \right)}^2{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2-2.2\cos \theta \left(\overrightarrow{b}·\overrightarrow{c}\right)$

Given that, $\left|\overrightarrow{a}\right|=\sqrt{2}, \left|\overrightarrow{b}\right|=1$ and $\left|\overrightarrow{c}\right|=2$

${\left(\sqrt{2}\right)}^2={\left(2\cos \theta \right)}^2+2^2-2.2\cos \theta \left(\overrightarrow{b}·\overrightarrow{c}\right)$

$\Rightarrow 2=4{\cos }^2\theta +4-4\cos \theta ·2\cos \theta$

$\left(\because \overrightarrow{b}·\overrightarrow{c}=\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos \theta \right)$

$\Rightarrow -2=-4{\cos }^2\theta$

$\Rightarrow {\cos }^2\theta =\frac{1}{2}$

$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \theta =\frac{\pi }{4}$

$\left(\because 0<\theta <\frac{\pi }{2}\right)$

Hence, $1+\tan \theta =2.$