This is called the re-transformation problem. I'm going to make your model a little simpler to talk about it:
$\ln{Y} = \beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3X_2^2 + \epsilon$
Now, that model does not make predictions for $Y$, it makes predictions for $\ln{Y}$. It is tempting to make predictions for $Y$ by just taking predictions for $\ln{Y}$ and exponentiating them like $\hat{Y}=\exp(\widehat{\ln{Y}})$. This is wrong (i.e. biased), though:
$\begin{align}
\ln{Y} &= \beta_0 + \beta_1X_1 + \beta_2X_2 +\beta_3X_2^2 + \epsilon\\
Y &= \exp(\beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3X_2^2)\exp(\epsilon)\\
\mathrm{E}\{ Y|X\} &= \exp(\beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3X_2^2)\,\mathrm{E}\{\exp(\epsilon)|X\}\\
\hat{\mathrm{E}}\{ Y|X\} &= \exp(\widehat{\ln{Y}})\,\mathrm{E}\{\exp(\epsilon)|X\}
\end{align}$
The best predictor of $Y$ is its expectation. If we could conclude that $\mathrm{E}\{\exp(\epsilon)|X\}=1$, then we could just exponentiate like you are suggesting above. But Jensen's inequality says that since $\mathrm{E}\{\epsilon|X\}=0$, it must be that $\mathrm{E}\{\exp(\epsilon)|X\}>1$. So, we have to use some kind of adjustment. The adjustment is called Duan's Smearing Estimator. It is just the sample mean of the exponentiated prediction errors (residuals) from the original model, $(1/N)\sum \exp(e_i)$. So the right way to re-transform from the log model back to predictions on Y is:
$$\hat{Y}_j = \exp(\widehat{\ln{Y}}_j) \cdot \frac{1}{N}\sum_{i=1}^N \exp(e_i)$$
To your questions. On the parameters, whether you need to re-transform depends on what you are trying to measure. The parameter $\beta_2$ measures the amount that $Y$ goes up (in percents) for a one unit increase in $X_1$. So, if $\beta_2=0.04$, that says that $Y$ goes up 4% for each one unit $X_1$ goes up. Similarly, for each unit $X_2$ goes up, $Y$ goes up $\beta_2+2\beta_3X_2$ percent.
If you want to measure the amount that $Y$ goes up in units when $X_2$ goes up by one unit, then you need to re-transform:
$\begin{align}
Y &= \exp(\beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3X_2^2)\exp(\epsilon)\\
\frac{\partial Y}{\partial X_1}
&= \exp(\beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3X_2^2)\exp(\epsilon)\beta_1\\
\hat{\mathrm{E}}\left\{ \frac{\partial Y}{\partial X_1} \right\}
&= \exp(\widehat{\ln{Y}}) \cdot \frac{1}{N}\sum \exp(e_i) \cdot \beta_1
\end{align}$
Notice that the answer depends on $\hat{Y}$, totally unlike "regular" regression. You should expect this, though. The model is non-linear, so the derivative depends on the point of evaluation. For your more complicated model, you have to be careful to apply the chain rule properly---that is, where I have $\beta_1$, you will have a complicated expression with $\beta$s and powers of your various $X$s and such.
For the confidence intervals, again, the question is what you are trying to measure. If you are happy with knowing how many percents $Y$ goes up when $X_1$ goes up by one, then the "regular" confidence intervals you get from the usual regression output are fine. If you want to measure the number of units that $Y$ goes up when $X_1$ goes up by one, then it's more complicated. Actually, it's very complicated in that case---you should use bootstrapping to do it. You can use something called the delta method, but it is a pain.
Root mean squared error of the prediction is easy to calculate, once you have re-transformed back to predicted $Y$:
$$\mathrm{RMSEP} = \sqrt {\frac{1}{N-1} \sum (\hat{Y}_i-Y_i)^2}$$
where $\hat{Y}_i$ comes from the formula above.
