Maybe it helps to do this for $n = 2$, as a start.
There are three key results that will help.
(1) If $X_1$ and $X_2$ have $E(X_1) = \mu_1$ and
$E(X_2) = \mu_2,$ then
$$E(a_1X_1 = a_2X_2) = a_1E(X_1) + a_2E(X_2) = a_1\mu_1 + a_2\mu_2.$$
For $\bar X = \frac 1 2 (X_1 + X_2),$ take $a_1 = a_2 = 1/2$ and
$\mu_1 = \mu_2 = \mu,$ to get $E(\bar X) = \mu.$ So the mean is right.
(2) Similarly, if $X_1$ and $X_2$ are independent and have $Var(X_1) = \sigma_1^2$ and
$Var(X_2) = \sigma_2^2,$ then
$$Var(a_1X_1 = a_2X_2) = a_1^2 Var(X_1) + a_2^2 Var(X_2) =
a_1^2\sigma_1^2 + a_2^2\sigma_2^2.$$
For $\bar X = \frac 1 2 (X_1 + X_2),$ take $a_1 = a_2 = 1/2$ and
$\sigma_1^2 = \sigma_2^2 = \sigma^2,$ to get $Var(\bar X) = \sigma^2/2.$ So the variance is right.
(3) If $X_1 \sim \mathsf{Norm}(\mu_1, \sigma_1)$ and independently
$X_2 \sim \mathsf{Norm}(\mu_2, \sigma_2),$ then you can use moment
generating functions to prove that $X_1 + X_2 \sim
\mathsf{Norm}\left(\mu_1 + \mu_2,\, \sqrt{\sigma_1^2 + \sigma_2^2}\right).$
So the normal distribution part is right. That is, the sum of two independent normal random variables is a normal
random variable; the means add and the variances add (not the standard deviations).
Finally, you have
$\bar X = \frac 1 2(X_1 + X_2) \sim \mathsf{Norm}(\mu, \sigma^2/2).$
So there is lot involved in the statement you posted. Without some careful
explanation before or after this statement pointing out connections to
the three results I have mentioned, this is a pretty big chunk to absorb
at first glance. But, taken separately, each part of the explanation is
simple enough.
I hope you can find statements in your text similar
to my (1), (2), and (3) and make sure you see how they apply. Maybe
some proofs are postponed until later, but make what linkages you can for now,
and revisit the statement later as necessary.
Your second question seems to be about estimating the population parameters $\mu$ and $\sigma$ from a large sample. For large $n,$ you have $\bar X_n$ very close to $\mu.$ So you want $\bar X_n$ to have a small variance in order to make $\bar X_n$ have good precision as an estimate of $\mu.$
Estimating the population variance $\sigma^2$ by the sample variance $S_n^2$ of a large sample is a separate issue. First, one can show thate $E(S_n^2) = \sigma^2.$ Second, one can show that $\frac{(n-1)S_n^2}{\sigma^2} \sim
\mathsf{Chisq}(df = n-1).$ Without going into details prematurely, this
implies that $S_n^2$ gets ever closer to $\sigma^2$ as $n$ increases.
I sampled $n=1000$ observations from $\mathsf{Norm}(\mu = 100, \sigma = 10)$ and got $S_{1000}^2 = 100.2097.$ Using the chi-squared distribution this gives
$(91.97, 109.61)$ as a 95% confidence interval for $\sigma^2$ and [taking
square roots] also $(9.59, 10.47)$ as a 95% CI for $\sigma.$
.
