Do i get the right MLE and 90% confidence interval of normal distribution? 

I think i do right in step1 above. But i wonder whether i get the right confidence interval of mu and sigma in step2?
 A: You inquire specifically about the confidence intervals. 
It is worthwhile to distinguish several cases:
(1) $\mu$ is unknown and $\sigma$ is known. We seek a CI for $\mu$. Then $\bar X$ is
the MLE of $\mu$ with 
$$\frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim Norm(0, 1).$$
Then a 95% CI for $\mu$ is $\bar X \pm 1.96\sigma/\sqrt{n}.$
(2) $\mu$ and $\sigma$ are both unknown. We seek a CI for $\mu$. 
Then $\bar X$ is the MLE of $\mu$ and 
$S = \sqrt{\frac{\sum_{i=1}^n(X_i - \bar X)^2}{n-1}}$
is the MLE of $\sigma.$ Then a 95% CI for $\mu$ is
$\bar X \pm t^*S/\sqrt{n},$ where $t^*$ cuts 2.5% from the
upper tail of Student's t distribution with $n - 1$ degrees
of freedom.
(3) $\sigma$ is unknown and $\mu$ is known. We seek a CI for $\sigma.$  Then $nV/\sigma^2 \sim Chisq(df = n),$
where $V = (1/n)\sum_{i=1}^n (X_i - \mu)^2.$
This relationship can be used to get a 95% CI for $\sigma^2.$
Take square roots of both endpoints to get a CI for $\sigma.$
(4) $\sigma$ and $\mu$ are both unknown. We seek a CI for $\sigma.$ 
Then $(n-1)S^2/\sigma^2 \sim Chisq(n-1).$ This relationship can be used to get a 95% CI for $\sigma^2.$
Take square roots of both endpoints to get a CI for $\sigma.$
(5)  $\sigma$ and $\mu$ are both unknown. We seek a 2-dimensional
confidence region for both parameters simultaneously.
Because $\bar X$ and $S^2$ (based on normal data) are independent random variables, we can use the ideas of (2) and (4) to get a
region in $(\mu, \sigma^2)$-space that is a 95% confidence
region for the two parameters simultaneously. Details are not
difficult, and are shown (among other places) in Mood/Graybill/Boes, 3e (1974) page 384. This region (bounded by a horizontal strip and a parabola) does not have minimum area for
a fixed confidence level. Roughly speaking, its area can be
slightly reduced by making a roughly elliptical region that
"rounds the corners" of the more elementary region--without
changing its coverage probability.
