(1) says that there is exactly one object (x) that satisfies unary predicate F.
(2) says that there exists some unary predicate (X) for which there is exactly one object (x) that satisfies it, and that object also happens to satisfy unary predicate F.

The sentences are not equivalent. In particular, the second statement does not say that only one object satisfies F. It could be the case that there is a different unary predicate G that is satisfied by exactly one object a, and that Fa is true but so is Fb for some b not equal to a.

What is the statement you are trying to encode? 'There is the F' doesn't mean anything to me, and hence cannot be rendered meaningfully in either natural language or formal logic.

Let's say 'The F is G' instead.

(3) (∃x)(Fx & (∀y)(Fy → x = y) & Gx)

I think that (3) completely express 'The F is G' in first-order logic. The question would be: What is the equivalent form to (3) in second-order logic?

I don't know what 'The F is G' means either. Can you give a concrete example, using natural language predicates that people would understand - like 'is a mammal' and 'is an animal', rather than letters like F and G?

The equivalent to (3) in second-order logic is just (3). Since it can be expressed in first-order logic, no translation is required.

(4) The king of France is bald.

I think that (4) can be expressed as (3) in first-order logic. Is it possible that (4) is expressed in second-order logic?

How about this:

(5) (∃X)(∃x)((Xx & (∀y)(Xy → x = y)) & Fx & (∀y)(Fy → x = y))

(4) The king of France is bald. — quine

Based on the thread title, I assume that you intend to be asserting here that the king of France exists and is bald, which would be consistent with (3). However, (4) is ambiguous enough in natural language that it could also be interpreted as asserting that any king of France is bald, regardless of whether the king of France (currently) exists.

(6) (∀x)(Fx → Gx)

Either way, I agree with @andrewk that there is no obvious reason why the expression in second-order logic would be any different, since you do not need to quantify over predicates.

Russell interprets (4) in expanded natural language as something like the following:

- There exists a person that is currently king of France
- If anybody is currently king of France then they are the same person as the one mentioned in the previous point
- that person is bald

This can be symbolised in First-Order Predicate Logic as

∃x (IsKingOfFrance(x) ∧ ∀y(IsKingOfFrance(y) → y=x) ∧ IsBald(x))

There is no need for second-order logic or set theory in this case.

Some statements are expressed in second-order logic even though they can be expressed in first-order logic. For example, 'A supervenes on B' can be enough translated into first-order logic, but Stanford Encyclopedia 'supervenience' entry states its formula by second-order logic. This means all first-order statements can be symbolized as second-order statements as all propositional calculuses can be expressed as predicate calculuses.