Given,

$\left|\left(\hat{a}+\hat{b}\right)+2\left(\hat{a}\times \hat{b}\right)\right|=2,\theta \in \left(0,\pi \right)$

Squaring both side we get,

$\left(\left(\hat{a}+\hat{b}\right)+2\left(\hat{a}\times \hat{b}\right)\right)·\left(\left(\hat{a}+\hat{b}\right)+2\left(\hat{a}\times \hat{b}\right)\right)=4$

${\left|\hat{a}+\hat{b}\right|}^2+4{\left|\left(\hat{a}\times \hat{b}\right)\right|}^2+0=4$

Let the angle be $\theta$ between $\hat{a}$ and $\hat{b}$

$\Rightarrow {\left|\hat{a}\right|}^2+{\left|\hat{b}\right|}^2+2\left|\hat{a}\right|\left|\hat{b}\right|\cos \theta +4\left(\left|\hat{a}\right|\left|\hat{b}\right|\sin \theta \right)=4$

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

$\Rightarrow 2+2\cos \theta -4{\cos }^2\theta =0$

Let $\cos \theta =t$ then

$\Rightarrow 2t^2-t-1=0$

$\Rightarrow 2t^2-2t+t-1=0$

$\Rightarrow 2t\left(t-1\right)+\left(t-1\right)=0$

$\left(2t+1\right)\left(t-1\right)=0$

So, $t=-\frac{1}{2}$ or $t=1$

Then, $\cos \theta =-\frac{1}{2}$ or $\cos \theta =1$ which is not possible as $0\in \left(0,\pi \right)$

So, $\boxed{\theta =\frac{2\pi }{3}}$

Now,

$S_1\Rightarrow 2\left|a\times b\right|=2\sin \left(\frac{2\pi }{3}\right)=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$

And $\left|\hat{a}-\hat{b}\right|=\sqrt{1+1-2\cos \left(\frac{2\pi }{3}\right)}$

$=\sqrt{2-2\times \left(-\frac{1}{2}\right)}$

$=\sqrt{3}$

$S_1$ is correct.

$S_2$ projection of $\hat{a}$ on $\left(\hat{a}+\hat{b}\right)$

$\Rightarrow \frac{\hat{a}.\left(\hat{a}+\hat{b}\right)}{\mid \hat{a}+\hat{b})}=\frac{1+\cos \left(\frac{2\pi }{3}\right)}{\sqrt{2+2\cos \frac{2\pi }{3}}}$

$\Rightarrow \frac{1-\frac{1}{2}}{\sqrt{1}}=\frac{1}{2}$

Which is true, So both are true.