= Solution
Assume $n\ge2$ and $\sigma^2>0$. Up to a constant, the <log-likelihood> of the <normal sample> is
$$
\ell(\mu,s)=-\frac n2\log s-\frac1{2s}\sum_i(X_i-\mu)^2,\qquad s=\sigma^2.
$$
For each $s$, the square is minimized at $\mu=\bar X$; optimizing the remaining one-variable likelihood gives the <maximum-likelihood estimators>
$$
\boxed{\widehat\mu=\bar X,\qquad
\widehat\sigma^2=\frac1n\sum_i(X_i-\bar X)^2.}
$$
They are the global maximum when the residual sum is positive, which occurs with probability one. On the exceptional all-equal sample the likelihood increases without bound as $s\downarrow0$.
To derive the distributions and independence, write $Z_i=(X_i-\mu)/\sigma$ and perform an orthogonal change of coordinates whose first row is $n^{-1/2}(1,\ldots,1)$. The transformed components $Z'_1,\ldots,Z'_n$ remain independent standard <normal random variables>, since their joint density depends only on the squared <Euclidean norm>. Then
$$
\widehat\mu=\mu+\frac{\sigma}{\sqrt n}Z'_1,
\qquad \frac{n\widehat\sigma^2}{\sigma^2}=\sum_{j=2}^n(Z'_j)^2.
$$
The first expression uses only $Z'_1$, the second only the other components. Consequently
$$
\boxed{\widehat\mu\sim N(\mu,\sigma^2/n),\qquad
n\widehat\sigma^2/\sigma^2\sim\chi^2_{n-1},\qquad
\widehat\mu\text{ and }\widehat\sigma^2\text{ are independent}.}
$$
This also exhibits why the maximum-likelihood variance uses divisor $n$, even though its chi-squared degrees of freedom are $n-1$.
The independent future observation satisfies $X_0-\bar X\sim N(0,\sigma^2(1+1/n))$. It is independent of the residual sum of squares, because both $X_0$ and $\bar X$ are independent of the residual vector. Combining this normal variable with the independent chi-squared variable gives
$$
T=\frac{X_0-\widehat\mu}{\widehat\sigma\sqrt{(n+1)/(n-1)}}\sim t_{n-1}.
$$
The event in the proposed <prediction interval> is equivalent to $|T|<\gamma\sqrt{(n-1)/n}$. By symmetry and continuity of <Student's t-distribution>, exact joint coverage $1-\alpha$ therefore requires
$$
\boxed{\gamma=\sqrt{\frac n{n-1}}\,t_{n-1,1-\alpha/2}.}
$$
This is the <normal prediction interval with maximum-likelihood variance>. The factor $\sqrt{n/(n-1)}$ cannot be dropped: the interval uses $\widehat\sigma$, not the unbiased residual standard deviation. Coverage here is over repeated joint sampling of the old and new observations, not a claim of parameter-free conditional coverage for every realized sample.
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