For clarification 1, evaluate the expectation and then use the property that Var[aX] = a^2*Var[X].
For clarification 2, you need to use the CDF result for e_it and standardize the expression so the distribution has mean 0 and variance 1.
Think of how you do this for a Normal distribution. Lets say you have X ~ N(mu,v^2). You can then say
P(X < x) = P(X - mu < x - mu) = P([X - mu]/v = (x - mu)/v = P(Z < (x - mu)/v) = P(Z < z).
Its exactly the same sort of idea.
Thanks chiro, looks like I made things harder than it seemed haha