Estimating confidence Interval for unknown Variance, Normal distribution I've been stuck with this question for a while:

I've learnt how to use t-distribution to estimate CIs for an unknown variance, but I'm unsure how that applies to this situation.
Any help would be greatly appreciated.
Thank you!

$X_1,\ldots,X_n$ are i.i.d. $\mathcal N(\mu,\sigma^2)$ random variables and $\bar X=n^{-1}\sum_{i=1}^nX_i$. The distributions and confidence intervals are as follow.
(a)
$$
\frac1{\sigma^2}\sum_{i=1}^n(X_i-\bar X)^2\sim\chi_{n-1}^2
$$
and
$$
\Pr\biggl(\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\chi_{n-1,\alpha/2}^2}\le\sigma^2\le\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\chi_{n-1,1-\alpha/2}^2}\biggr)=1-\alpha.
$$
(b)
$$
\frac1{\sigma^2}\sum_{i=1}^n(X_i-\mu)^2\sim\chi_n^2
$$
and
$$
\Pr\biggl(\frac{\sum_{i=1}^n(X_i-\mu)^2}{\chi_{n,\alpha/2}^2}\le\sigma^2\le\frac{\sum_{i=1}^n(X_i-\mu)^2}{\chi_{n,1-\alpha/2}^2}\biggr)=1-\alpha.
$$
(c)
$$
\frac{n(\bar X-\mu)^2}{\sigma^2}\sim\chi_1^2
$$
and
$$
\Pr\biggl(\frac{n(\bar X-\mu)^2}{\chi_{1,\alpha/2}^2}\le\sigma^2\le\frac{n(\bar X-\mu)^2}{\chi_{1,1-\alpha/2}^2}\biggr)=1-\alpha.
$$
I hope this helps.
