The question requires to show that $\sqrt{2\chi^2_n}-\sqrt{2n}$ converges in distribution to $N(0,1)$ as $n \rightarrow \infty$, which I dont know how to proceed.
The question also has a first part asking to show $\frac{\chi^2_n-n}{\sqrt{2n}}$ converges in distribution to $N(0,1)$, and it's straightforward by CLT. I am thinking if there is any connection between the two parts, however still have no idea.
Can anyone help me ? Thanks in advance.
You can use the univariate delta method: link. First we start with a situation where the CLT holds. In your case you can represent a $\chi^2_n$ variable as $\sum_{i=1}^{n} Z_i^2 $ with ${Z_i}$ iid $N(0,1).$ Then $\mu:=E(Z_i^2)=1$ and $\sigma^2:=Var(Z_i^2)=2.$ Let $B_n= \frac1n \sum_{i=1}^{n} Z_i^2.$ Then the CLT shows $ \sqrt n{[B_n-\mu]}\to N(0,\sigma^2)$ in distribution.
Now we use the transformation $g(x)=\sqrt{2x}$ on the statistic $B_n $ and investigate the asymptotic distribution of $g(B_n).$ The univariate delta method says we can conclude: $$ \sqrt{n}[g(B_n)-g(\mu)]\to N(0,\sigma^2[g^{'}(\mu)] ^2 )$$
Now $g^{'}(x)=1/\sqrt{2x} $ so we have $$ \sqrt{n}[\sqrt{2\chi_n^2/n}-\sqrt{2}]\to N(0,1 )$$ and that is what you wanted.