Qun 3
(I notice you have not eliminated the quantifiers, as asked in the question).
Also, P(1) being true, doesn't mean that ~P(1) v Q(y) is false (disjunction). Q(y) could be true.
In fact a direct reading of the original statement shows that it is true.
[MATH]\forall x \exists y\left(P\left(x\right)\rightarrow Q\left(y\right)\right)[/MATH]For x=-1, P(-1) is false therefore the implication is True for every value of y (so certainly there exists one).
For x=1, P(1) is true therefore for the implication to be true we need to have a value of y for which Q(y) is true. There does exist such a y, namely y=3.
Therefore for each value of x there does exist a y for which P(x) -> Q(y)
Qun 4
The first bracket of your answer should be a conjunction rather than a disjunction and then this all can be simplified further.
