I think that you are asking for :
how to prove that (∃y)(∀x)Fxy is not a logical consequence [see this post for the definition] of (∀x)(∃y)Fxy.
If so, in order to prove it, we have to find a counter-example, i.e. an interpretation such that (∀x)(∃y)Fxy is true while (∃y)(∀x)Fxy is false.
The "standard" counter-example is found assuming as domain for the interpretation the set N of natural numbers and with the relation < ("less-then") as interpretation for the binary predicate symbol F.
We have that in N it is true that :
(∀x)(∃y)(x < y)
because for every natural number n it is enough to choose n+1 and we have n < n+1.
(∃y)(∀x)(x < y)
is false, because there is no number which is greater than all other numbers.
Regarding the second problem, it amounts to show that :
(∀x)(Fx → Gx) and ¬(∃x)(Fx & Gx) are simultaneously satsfiable.
We can show it applying some simple transformations.
1) A → B is equivalent to ¬A ∨ B
2) ¬(A & B) is equivalent to ¬A ∨ ¬B [De Morgan]
3) ¬(∃x)A is equivalent to (∀x)¬A.
Consider now : (∀x)(Fx → Gx); by 1) it is equivalent to : (∀x)(¬Fx ∨ Gx).
Consider : ¬(∃x)(Fx & Gx); by 3) it is equivalent to : (∀x)¬(Fx & Gx) and by 2) to : (∀x)(¬Fx ∨ ¬Gx).
Thus the problem is equivalento to show that :
(∀x)(¬Fx ∨ Gx) and (∀x)(¬Fx ∨ ¬Gx) are simultaneously satsfiable.
We can prove this assuming a domain with only one black ball and interpret the two predicate symbols Fx and Gx as "x is white" and "x is square" respectively.
With this interpretation, the first formula : (∀x)(¬Fx ∨ Gx) means :
"all objects in the domain are not-white and square"
while the second formula : (∀x)(¬Fx ∨ ¬Gx) means :
"all objects in the domain are not-white and not-square".
Due to the fact that in the domain there is only one black ball, i.e. a not-white ball, both disjunctions are satisfied simultaneously.