Predicting $y$ when response variable is $\ln(y)$
You do not find the same answer because of what I suspect to be a typographical error, which would thus be the main reason of your problem: $x_{03}$ would be set to $80$, not $8$. Another possibility, if you keep $x_{03}=8$, is an error in the second estimated coefficient, say, $\hat{\beta}_2 = -0.1$ instead of $-0.01$.

Anyway, one of these modifications solves everything and yields the same result as the solution to this exercise.

Considering this change, with $t_{\alpha/2}=1.96476138969835$, one gets

Method 1

$\text{CI}(E[y_0|x_0])=[e^{6.43618291164626},e^{6.49755798189177}]=[624.020307335178,663.519326788772]$
(the given solution to this exercise)

or Method 2
(as stated In Wooldridge's Intro to Econometrics, in page 212) if we're sure the error terms are normal (and one is extremely lucky)

$$\text{CI}(E[y_0|x_0])=\left[e^{s^2/2}624.0203,e^{s^2/2}663.5193 \right] = \left[ 624.0960,663.6002 \right] $$

however

the method 2 is very unlikely to be correct, since as you mention in your question [...] the (underestimation) correction should not be used to the CI but only for the point estimation.

Why ? I would say because of the dependency betweeen the two terms, knowing the expectations of $e^{s^2/2}$ on the one hand and $\hat{y_0}$ on the other hand does not mean one knows the one of $e^{\frac{s^2}{2} + \hat{\ln(y_0)}}$.