You shouldn't try to just randomly rephrase the sentence, you need to find its logical form. That is, you need to separate out all the predicates and express the sentence with quantifying phrases over conjunctions, disjunctions, conditionals and negations.

Your formalisation of 1 is correct.

2. If the product of two integers is even, then at least one of them is even.

First break down the clauses of the sentence.
If [the product of two integers is even] then [at least one of them is even].

We know from the 'if... then' bit of the sentence that we have a conditional, so there's a starting point. Before we introduce the quantifiers we need to figure out what we are actually talking about. In the first clause we a clearly talking about the product of two integers, and in the second clause we are talking about the two integers and not the product, so we know that our formalisation should contain three variables. We can also see that this is a universal claim because it's making a generalisation about all integers, so these three variables should be universally quantified.

Now, we need to express each clause in terms of the predicates we have been given. M: [1] * [2]= [3] is the only way to express multiplication, and we have an 'is even' predicate so:

For all x y and z[If [x * y = z] and [z is even]] then [at least one of them is even]]

Onto the second clause, 'at least one of them' might tempt you into introducing an existential quantifier, but we can see that 'them' refers back to x and y, and they have already been quantified over the whole domain. We are saying that x or y is even, so it's disjunction.

For all x y and z[If [x * y = z] and [z is even]] then [x or y is even]]

But we can delve a little deeper into the logical structure of the second clause.

For all x y and z[[If [x * y = z] and [z is even]] then [[x is even] or [y is even]]]

Now that the logical structure is fully revealed all we have to do is replace the clauses with predicate letters, the quantifying phrases with quantifiers and the logical connectives with logical symbols, so we get:

∀x∀y∀z((Mxyz & Ez) -> (Ex v Ey))

3. The product of two even integers is always a multiple of four.

Again, this is a universal claim about all even integers an their products, and it also takes the form of an if then statement:
For all xyz [If [z is the product of two even integers x and y] then [z is a multiple of four]]

Now, to put the first clause into it's proper logical form in much the same way as we did in last example:

For all xyz[If [[x * y = z] and [x is even] and [y is even]] then [z is a multiple of four]]

Now the question is how do we express "is a multiple of 4" with the predicates we have been given. It seems to me that if a number, z is a multiple of 4 then there exists another number, v, such that v * 4 = z. So not we have:
For all xyz[If [[x * y = z] and [x is even] and [y is even]] then [there exists a v such that [v * 4 = z]]]

Now that the full logical structure is revealed we can simply plug in the symbols and predicate letters:
∀x∀y∀z((Mxyz & Ex & Ey) -> ∃v(Mdvz))

4. Every integer greater than one is evenly divisible by some prime.
Another universal claim about all integers greater than one, and again it takes the form of an if then statement:

For all x[if [x is greater than 1] then [there is a prime number by which x is divisible]]

Now we can formalise the first clause in terms of predicates we've been given very easily. The second clause is an existential claim, we are saying that there exists at least one prime number, z, that divides into x. The question seems to be what do we mean by is "divisible by". Well if x is divisible by z then there is some number such that y * z = x. Therefore:

For all x[if [1 < x] then [there exists a y and there exists a z such that [y * z = x] and [[z is prime] or [y is prime]]]

Again we translate this into the logical calculus:

∀x((Lax) -> (∃y∃z((Myzx) & (Py v Pz))))

5. There is no greatest prime.

The sentence as a whole is the negation of an existential claim so we can start off with:

It is not the case that [there exists an x such that[x is the greatest prime number]]

At the moment there is only one clause to this sentence, but it should be possible to smear out all of the predicates to reveal the logical structure:

It is not the case that [there exists an x such that[[x is a prime number] & [x is larger than all prime numbers]]]

Now, the second clause, once again, is in the form of an if... then conditional:

If [[y is a prime number] then [x (here we refer back to the first clause) is larger than y]

As we are talking about all prime numbers, we need a universal quantifier. You can think of the resulting sentence as saying something about the way in which every object in the domain is related to a particular object is the domain. So now we have:

It is not the case that [there exists an x such that[[x is a prime number] and [for all y[if [x is a prime number] then [x is larger than y]]]]]

Before we translate into logical syntax it seems implicit in 'the greatest prime number' that there should only be one of them. We express this as: for all y [x=y].

The fully analysed logical structure:

It is not the case that [there exists an x such that[[x is a prime number] and for all z[x = z] and [for all y[if [y is a prime number] then [x is larger than y]]]]]

In the logical langauge:
~∃x(Px & ∀zIxz & ∀y(Py -> Lyx))

You might notice that whenever we use the universal quantifier we end up with a conditional. Generally speaking, this is how we formalise universal claims. What we we are effectively doing with the antecedent of the conditional is restricting our claim to a certain class of objects within the domain. However, if we want to ascribe a certain predicate to all objects in the domain, we needn't use a conditional e.g.
∀yNy
N: ... is a number