Let $z=x+i y$,  $\bar{z}=x-i y$

Now, $z=1-\bar{z}$

$\Rightarrow \,\, x+i y=1-(x-i y)$

$\Rightarrow \,\, 2 x=1 \Rightarrow x=\frac{1}{2}$

Now, $|z|=1 \Rightarrow x^{2}+y^{2}=1 \Rightarrow y^{2}=i-x^{2}$

$\Rightarrow  \,y=\pm \frac{\sqrt{3}}{2}$

Now, $\tan \theta =\frac{y}{x}$ ( $\theta $ is the argument) $=\frac{\sqrt{3}}{2} \div \frac{1}{2}$

( $+\,ve$ since only principal argument)

$=\sqrt{3}$

$\Rightarrow \theta=\tan ^{-1} \sqrt{3}=\frac{\pi}{3}$

Hence, $z$ is not a real number

So, statement $-1$ is false and $2$ is true.