a) Without loss of generality, we can consider this problem for nonnegative integers. We know that a perfect square gives the remainder $0$ or $1$ when divided by $3$, so all numbers from $A$ when divided by $3$ give the remainder $3$ or $2$.

Assume that there exists a pair $(x_0, y_0)$ such that $2x_0^2 + 3y_0^2 \in A$ is a perfect square and such that $x_0 + y_0$ is minimal. Then this number should be divisible by $3$, so $x_0$ is divisible by $3$. Put $x_0 = 3x_1$ where $x_1 \in \mathbb{Z}$.

Thus $18x_1^2 + 3y_0^2 = 3(6x_1^2 + y_0^2)$ is a perfect square. This number is divisible by $3$, so it should be divisible by $9$. So $6x_1^2 + y_0^2$ is divisible by $3$. From which we have $y_0$ is divisible by $3$. Put $y_0 = 3y_1$ where $y_1$ is an integer.

Putting back in original expression $2x_0^2 + 3y_0^2$, we have $18x_1^2 + 27y_1^2 = 9(2x_1^2 + 3y_1^2)$ is a perfect square. From which follows that $2x_1^2 + 3y_1^2$ is a perfect square.

However this contradicts the definition of the pair $(x_0, y_0)$ due to $x_1 + y_1 < x_0 + y_0$ (the equality does not hold because at least one of two numbers is positive). Thus, in $A$ does not exist a perfect square.

b.
Firstly we will prove that the product of any three numbers from $A$ also belongs to $A$. (*)

Indeed, consider numbers $2a^2 + 3b^2$, $2c^2 + 3d^2$, $2e^2 + 3f^2$ from $A$. We have
$$(2a^2 + 3b^2)(2c^2 + 3d^2)(2e^2 + 3f^2) = ((2ac - 3bd)^2 + 6(ad + bc)^2)(2e^2 + 3f^2).$$
Putting $m = 2ac - 3bd$, $n = ad + bc$, we have
$$(m^2 + 6n^2)(2e^2 + 3f^2) = 2(me + 3nf)^2 + 3(mf - 2ne)^2 \in A.$$

Next, we will prove the statement of the problem by induction.
- For $n = 1, 3$, from (*) the statement is true.
- Assume that the statement is true for $n = 2k-1 \ge 3$, i.e., the product of any $2k-1$ numbers from $A$ also belongs to $A$.

Consider any $2k+1$ numbers from $A$. Choose from them $2k-1$ numbers. Then, by induction hypothesis, the product of these numbers belongs to $A$. Then this product together with two remaining numbers forms a triple of numbers from $A$. By (*) their product belongs to $A$, which means the product of any $2k+1$ numbers from $A$ belongs to $A$. Thus, the statement holds for $n = 2k+1$.

By induction principle, we have the required statement.

Now, apply result of part a for the product, we have done.