Use the t-test to test
$H_0 : \beta_1 = 0$
$H_1 : \beta_1 \neq 0$
at level significance $5 \%$. Give test statistic and conclude within the
context of the study
Given
$SSREG = 36460$
$s = 365$
$\hat\beta_1 = -5.98$
$n = 49$
attempt:
$t^* = \frac{\hat\beta_1}{s(\hat\beta_1)}$
$s(\hat\beta_1) = \sqrt{\frac{s^2}{\sum_{i=1}^{n} (X_i - \bar{X})^2}}$
$\sum_{i=1}^{n} (X_i - \bar{X})^2 = \frac{SSREG}{\hat\beta_1^2} = 1019.5$
so then
$s(\hat\beta_1) = \sqrt{\frac{s^2}{\sum_{i=1}^{n} (X_i - \bar{X})^2}} = \sqrt{\frac{365^2}{1019.5}} = 11.4$
and finally
$t^* = \frac{(-5.98)}{11.4} = -0.52$
I have to show $|t^*| < t_{1 - \frac{\alpha}{2}, n - 2}$ or
$|t^*| \geq t_{1 - \frac{\alpha}{2}, n - 2}$
but what is $\alpha$ equal to?
Your previous approach is right. $\alpha$ is the significance level. This is the probability of the study rejecting the null hypothesis, given that the null hypothesis were true. In your case the null hypothesis is $H_0: \beta_1=0$ and the alternative hypothesis is $H_A: \beta_1\neq 0$.
The significance level is $5\%=0.05$. That means that $1-\frac{\alpha}{2}=0.975$
Decision rule
If $|t^*|=0.52\geq t_{0.975,47}$ you reject the null hypothesis, otherwise you don't.
The value of $t_{0.975,47}$ can be looked up in a table of the t-distribution, for instance here. Your test is a two-tailed test and the number of degrees of freedom is $47$.