i am wondering universal quantifier combined with implicationc

Question

Suppose that the domain of the propositional function P (x) consists of the integers 1, 2, 3, 4, and 5. Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions.

e) ∀x((x ≠ 3) → P (x)) ∨ ∃x¬P (x)

but i have a confused that

if x is 3 then (x ≠ 3) is false which means F→ P (x),whatever P(3) is true or false,the statment ∀x((3 ≠ 3) → P (3)) will also be true.Therefore,P(3) is included

if x is 1,2,4,5,then (x ≠ 3) is true which means T→ P (1), T→ P (2),T→ P (4),T→ P (5) is true.

∃x¬P (x) will be (¬P(1)∨¬P(2)∨¬P(3)∨¬P(4)∨¬P(5)).

so the final answer will be (P(1)∧P(2)∧P(3)∧P(4)∧P(5))∨(¬P(1)∨¬P(2)∨¬P(3)∨¬P(4)∨¬P(5)).

However,the actual answer is (P(1)∧P(2)∧P(4)∧P(5))∨(¬P(1)∨¬P(2)∨¬P(3)∨¬P(4)∨¬P(5)).

Answer 1

if x is 3 then (x ≠ 3) is false which means F→ P (x),whatever P(3) is true or false, the statment ∀x((3 ≠ 3) → P (3)) will also be true. Therefore, P(3) is included

No, you must include all the evaluations of $P$ that make the conditional true; that means that $(P(3)\vee\neg P(3))$ is included in the leftmost conjunction. Which is a tautology so...

Answer 2

Your disagreement with the published answer is whether $P(3)$ should be included in the first disjucnt. You have correctly noted that if $x$ is $3$ the $\forall x$ term is true regardless of whether $P(3)$ is true or not, so there is no reason to require it to be true. You need all the others to make the $\forall x$ term true, which is why they are all there.