This is probably too late, but here are the solutions anyway...
# 1
This is one tricky derivation, so you need to be a little creative with this one.
Note: each premise is used and when a new premise is derived from, I separated it to keep things looking clean.
1. (p <--> q) -> s
2. ~(~r -> t)
3. ~q v ~s Therefore: (t v p) -> (~t & ~ q)
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4. ~(~~r v t) 2 Material Implication
5. ~(r v t) 4 Double Negation
6. ~r & ~t 5 De Morgan's Law
7. ~t & ~r 6 Commutation
8. ~t 7 Simplification
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9. ~s v ~q 3 Commutation
10. s -> ~q 9 Material Implication
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11. (p <--> q) -> ~q 1,10 Hypothetical Syllogism
12. [(p & q) v (~p & ~q) ] -> ~q 11 Material Equivalence
13. ~[(p & q) v (~p & ~q) ] v ~q 12 Material Implication
14. [~(p & q) & ~(~p & ~q) ] v ~q 13 De Morgan's Law
15. ~q v [~(p & q) & ~(~p & ~q) ] 14 Commutation
16. [~q v (~p v ~q)] & [~q v (p v q) ] 15 Distribution
17. ~q v (~q v ~p) 16 Simplification
18. (~q v ~q) v ~p 17 Association
19. ~q v ~p 18 Tautology
20. ~p v ~q 19 Commutation
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21. ~t & (~p v ~q) 8,20 Conjunction
22. (~t & ~p) v (~t & ~q) 21 Distribution
23. ~(t v p) v (~t & ~q) 22 De Morgan's Law
24. (t v p) -> (~t & ~q) 23 Material Implication
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# 2
This derivation mainly involves isolating atomic components through modus ponens and a disjunction syllogism.
1. p v (q & r)
2. ~r
3. p -> (s -> ~t) Therefore: ~(s & t)
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4. (p v q) & (p v r) 1 Distribution
5. (p v r) & (p v q) 4 Commutation
6. p v r 5 Simplification
7. p 6,2 Disjunctive Syllogism
8. s -> ~t 3,7 Modus Ponens
9. ~s v ~t 8 Material Implication
10. ~(s & t) 9 De Morgan's Law