For Confidence interval, given $x*$, we have $E(Y|x_*)=\beta_0+\beta_1x_*$ Hence $\mathrm{Var} (Y_*)=\mathrm{Var} \{\hat\beta_0+\hat\beta_1x_*\}$ Using least squares estimators.
Hence we have $\hat\beta_0+\hat\beta_1x_*\ \pm t(n-2)\sqrt{\hat\sigma^2(1/n+\frac{(x_*-\bar x)^2}{S_{xx}})}$
or correct without sigma hat? $$\pm z(\alpha)\sqrt{\sigma^2(1/n+\frac{(x_*-\bar x)^2}{S_{xx}})}$$
Now to my question, regarding future response $Y_*=\beta_0+\beta_1x_*+\epsilon$ where $\epsilon \sim N(0,\sigma^2)$
For prediction interval, I have to consider: $$\hat\beta_0+\hat\beta_1x_*-Y_*=\hat\beta_0+\hat\beta_1x_*-\beta_0-\beta_1x_*-\epsilon$$
The variance is given by:
$$\mathrm{Var}\{\hat\beta_0+\hat\beta_1x_*-\beta_0-\beta_1x_*-\epsilon\}=\sqrt{\hat\sigma^2(1/n+\frac{(x_*-\bar x)^2}{S_{xx}})}-\mathrm{Var}\{\beta_0+\beta_1x_*+\epsilon\}$$
But $\mathrm{Var}\{\beta_0+\beta_1x_*+\epsilon\}=\mathrm{Var}\beta_0+\mathrm{Var}(\beta_1X_*)+\sigma^2=0+0+\sigma^2$
Hence we have our prediction interval:
$$\hat\beta_0+\hat\beta_1x_*\pm \sqrt{\hat\sigma^2(1/n+\frac{(x_*-\bar x)^2}{S_{xx}}-1)}$$ but my book says it should be $+1$
Edit added screenshot"
This stems from the property of variance that $\text{var}(-aX)=(-a)^2\text{var}(X)$, namely $$ \text{var}(\hat{Y}_i- Y_i)=\text{var}(\hat{Y}_i) +(-1)^2\text{var}(Y_i)=\sigma^2(1/n+(x_i - \bar{x})^2/S_{xx})+\sigma^2, $$ there is no covariance because $\hat{Y}_i$ and $Y_i$ are independent, so taking $\sigma^2$ as the common factor, you get
$$ \text{var}(\hat{Y}_i- Y_i)=\sigma^2(1/n+(x_i - \bar{x})^2/S_{xx}+1). $$