Variance and symmetry of Normal distribution

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I'm doing an exercise from a book and don't quite understand the parts I marked in blue (see pic below). Why is symmetry of the Normal invoked out of the blue? Also, why is $E(X-Y)^2=Var(X-Y)$ and not $E(X-Y)^2=E(X^2)-2E(XY)+E(Y^2)$? Furthermore, since $M = max(X,Y)$ and $L=min(X,Y)$ how is it that $E(X-Y)^2=E(M-L)^2$ (which would imply $M=X$ and $L=Y$)?

I've been thinking about this for over half an hour and am still completely puzzled. I feel like I'm missing something very basic and someone just needs to point it out and then the "how stupid I am" moment will follow.

ex53

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(2) $X, Y \sim N(0,1) \Rightarrow E[X]=E[Y]=0,\; V[X]=V[Y]=1$. Hence $E[X-Y]=E[X]-E[Y]=0$, by linearity of expectation. Also, by definition of variance, $V(X-Y)=E[(X-Y)^2]-(E[X-Y])^2=E[(X-Y)^2]-0^2=E[(X-Y)^2]$

(3) By definition of M & L, $M-L=|X-Y|\Rightarrow (M-L)^2=|X-Y|^2=(X-Y)^2\Rightarrow E[(M-L)^2]=E[(X-Y)^2]$ (which off course does not imply $M=X$ and $L=Y$)