$\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$Let $\al=\beta=1$. Let $Z_n:=\sqrt n\,X_n$, so that $Z_n\to Z\sim N(0,1)$ in distribution. Then for real $\ep>0$ $$ P\Big(\frac{X_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big) =P\Big(|Z_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n}}\,\Big) \\ \to P\Big(|Z|>\sqrt{\frac1{\ep^{-2}-1}}\,\Big) \ge 1-\de $$ if $\ep>0$ is small enough, depending on $\de$. $\quad\Box$

$\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$Let $\al=\beta=1$. Let $Z_n:=\sqrt n\,X_n$, so that $Z_n\to Z\sim N(0,1)$ in distribution. Then for real $\ep>0$ $$ P\Big(\frac{X_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big) =P\Big(|Z_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n}}\,\Big) \\ \to P\Big(|Z|>\sqrt{\frac1{\ep^{-2}-1}}\,\Big) \ge 1-\de $$ if $\ep>0$ is small enough, depending on $\de$. $\quad\Box$

To address a comment by the OP, alternatively we can take any real $\be>1$ and then $$ P\Big(\frac{X_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big) =P\Big(|Z_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n^\be}}\,\Big) \\ \to P(|Z|>0)=1\ge 1-\de. $$

$\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$Let $\al=\beta=1$. Let $Z_n:=\sqrt n\,X_n$, so that $Z_n\to Z\sim N(0,1)$ in distribution. Then for real $\ep>0$ $$ P\Big(\frac{X_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big) =P\Big(|Z_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n}}\,\Big) \\ \to P\Big(|Z|>\sqrt{\frac1{\ep^{-2}-1}}\,\Big) \ge 1-\de $$ if $\ep>0$ is small enough. $\quad\Box$

$\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$Let $\al=\beta=1$. Let $Z_n:=\sqrt n\,X_n$, so that $Z_n\to Z\sim N(0,1)$ in distribution. Then for real $\ep>0$ $$ P\Big(\frac{X_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big) =P\Big(|Z_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n}}\,\Big) \\ \to P\Big(|Z|>\sqrt{\frac1{\ep^{-2}-1}}\,\Big) \ge 1-\de $$ if $\ep>0$ is small enough, depending on $\de$. $\quad\Box$