Unfortunately, I don't know your background and don't know if you've studied gaussian vectors...
Question 1) is correct. More beautifully, you could say that X and -Y have the same distribution and hence the result.
Question 2) is T. Indeed, X and Y being independent normal rv's, (X,Y) is a Gaussian vector (any linear combination of X and Y has a normal distribution). So (X+Y,X-Y) is a Gaussian vector too, because any linear combination of X+Y and X-Y will be a linear combination of X and Y.
And if (Z,T) is a Gaussian vector, there is an equivalence between cov(Z,T)=0 and independence between Z and T. So just check here that cov(X+Y,X-Y)=0 and you'll have the result.
Given the following questions, try to find the joint pdf of (X+Y,X-Y) by any means you may have learnt.
Both 3) and 4) are T. Use characteristic functions for instance.
6) use the fact that X and -X are identically distributed and Y and -Y also are : P(X>Y)=P(-X>-Y)
7) cov(X,X+Y)= ? if it's not 0, then it can't be independent
8) is it an almost sure equality ? It's false since X and Y are continuous rv's and thus it is not possible, almost surely, for them (and for X²+Y² as well) to take a single value