Estimators of population parameters

1. "Let ${X_1, . . . , X_n}$ be a random sample from a population with mean $\mu = E(X_i)$. Find an estimator of $\mu$."

For this question, I was wondering why did they specifically say $\mu = E(X_i)$, and also not include what the variance is?

Is this because usually, if ${X_1, ..., X_n}$ is a random sample with a mean of $\mu$ and variance of ${\sigma}^2$, we assume each object from the population i.e. each $(X_i)$ is independent and identically distributed (as far as I have currently learnt). So, in this case - the questioner doesn't want us to assume that they are independent and identically distributed? The question goes on to talk about the sample mean $\bar{X}$ being a natural estimator for $\mu$.

2. "The variance of an estimator, denoted $Var(\hat{\theta})$, is obtained directly from the estimator's sampling distribution.
For the sample mean, $\bar{X}$, we have $Var(\bar{X}) = \frac{\sigma^2}{n}$."

Is it correct that this is only for: a population of a random variable X which is normally distributed, or approximately (for nearly every population of a random variable X which has a distribution with a finite variance) by the Central Limit Theorem? I was just wondering, since it wasn't specified and just assumed to be true.

3. Suppose that $\hat{\theta}$ is an estimator of the parameter, $\theta$. Then, is $\hat{\theta} - \theta$ the error of the estimator, and $|\hat{\theta} - \theta|$ the absolute error of the estimator?
If so, is the bias of an estimator: $E(\hat{\theta}) - \theta = E(\hat{\theta}) - E(\theta)$ (since $\theta$ is a constant), = $E(\hat{\theta} - \theta) =$ the expected value of the error of the estimator?

5. If $E(\hat{\theta^2}) < \infty$, it holds that: $MSE(\hat{\theta}) = Var(\hat{\theta}) + [Bias(\hat{\theta})]^2$ where $Bias(\hat{\theta}) = E(\hat{\theta}) - \theta$.
Why is it a requirement that $E(\hat{\theta}^2) < \infty$?
Is it because: $Var(\hat{\theta}) = E(\hat{\theta}^2) + (E(\hat{\theta}))^2$ and hence, if it is less than infinity, then the MSE is less than infinity, which I suppose might be a requirement that it is finite?

6. "Intuitively, MAD is a more appropriate measure for the error in estimation. However, it is technically less convenient since the function h(x) = |x| is not differentiable at x = 0."
How does this link? Is it that if we are differentiating to find the maximum value of MAD, we cannot do so?

Why should they? You're interested in estimating the population mean - you can do this without any mention of any other population parameter.

No. Provided the population is of infinite size, the very notion of "random sample" implies independent and identically distributed. I think you're simply over-reading things here - the question doesn't mention something here because it's irrelevant.

No. This is true in general.

Useful material here.

Well, how are you going to deal with infinite quantities?