First of all, can I first change
~A&~B by using the De Morgan rules?
Well sure, you could use derived rules, however the most convincing natural deduction argument for this commutivity would use the fundamental rules of inference; thus showing that commutivity follows directly from the introduction rules having left and right side versions.
After all, if you used deMorgan's, you would then commute a conjunction before using deMorgan's again. So since you do have to justify commutation either way, then you may as well do so without using any derived rules.
1|_ ~(A v B) : Premise
2| |_ B v A : Assumption
3| | |_ B : Assumption
4| | | A v B : Disjunction Introduction (3, Left)
| | +
5| | |_ A : Assumption
6| | | A v B : Disjunction Introduction (5, Right)
7| | A v B : Disjunction Elimination (2,3-4,5-6)
8| | # : Negation Elimination (1,7)
9| ~(B v A) : Negation Introduction (2-8)
And the second is:
I have to derive that this is a contradiction.
It seems hard because the entire argument is negated. I am not sure what to assume first.
Use the Law of Excluded Middle, assume A or ~A must be the case. To eliminate that disjunction to something usable, you have to introduce a disjunction and a negation of a conjunction.
0| A v ~A : Axiom (L.E.M.)
1|_ ~(A v ~(A & B)) : Premise
2| |_ A : Assumption
3| | A v ~(A & B) : Disjunction Introduction (2, R)
4| |_ ~A : Assumption
5| | |_ A & B : Assumption
6| | | A : Conjunction Elimination (5, R)
7| | | # : Negation Elimination (4,6)
8| | ~(A & B) : Negation Introduction (5-7)
9| | A v ~(A & B) : Disjunction Introduction (8, L)
10| A v ~(A & B) : Disjunction Elimination (0,2-3,4-9)
11| # : Negation Elimination (1,10)