I don't understand why $Var(X) = E((X-\mu)^2)$.
It's defined as the "expected value of the square of the deviation of $X$ from the mean" but I don't understand why it couldn't be $E(X-\mu)$ as that seems more intuitive for "deviation from the mean".
Is the purpose of the squaring to make deviations positive? Because if so, why not just $E(|X-\mu|)$ instead? What's the point of the squaring?
The mean absolute deviation (wrt to the mean) $E(\vert X - \mu \vert)$ is an alternative index of variability. A variant is to look at the mean absolute deviation wrt to the median $m$, because it can be shown that the mean absolute deviation $E(\vert X - a \vert)$ wrt to a value $a$ is minimised when $a=m$.
The variance is usually preferred to the mean absolute deviation for a few reasons. A modelling is that the quadratic term penalises large deviations more than small deviations (presumably, large deviations are worse). Another one is that the square deviation is differentiable and hence easier to handle.
You can learn more about this at https://stats.stackexchange.com/questions/118/why-square-the-difference-instead-of-taking-the-absolute-value-in-standard-devia