Are the following two formulas in first order logic equivalent?
$$¬\exists_x(P(x) \land Q(x)) \tag{1}$$
$$¬\exists_x(P(x)) \land Q(x) \tag{2}$$
Would the difference mean something like:
$$'\text{There is no P that is Q}'\tag{1}$$
$$\text{vs.}$$
$$'\text{There is no P, and that non-P is Q}'\tag{2}$$
A reference to a textbook would also be appreciated.
The first translation you did is correct, sencond is not.
And Yes, $(1)(2)$ are not equivalent. \begin{align} '\text{There is no P that is Q}' \equiv&\forall_x(P(x)\to\neg Q(x))\tag*{$\color{green}{\boxed{\unicode{x2714}}}$}\\ \equiv&\forall_x(\neg P(x)\lor\neg Q(x))\\ \equiv&\neg\exists_x (P(x) \land Q(x))\tag{1}\\\\ '\text{There is no P, and that non-P is Q}'\equiv&\forall_x(\neg P(x))\land\forall_x(\neg P(x)\to Q(x))\tag*{$\color{red}{\boxed{\unicode{x2718}}}$}\\ \equiv&\boxed{\forall_x(\neg P(x)\land Q(x))}\\ \not\equiv& \neg\exists_x(P(x)) \land Q(x)\tag{2}\\ \\ \text{Note two x are not in Scope of same quantifier}~~~~&\\ \text{To avoid confusing, we denote second x as y}~~~~\\ '\text{There is no P, and that y is Q}'\equiv&\boxed{\forall_x(\neg P(x))\land Q(y)}\tag*{$\color{green}{\boxed{\unicode{x2714}}}$}\\ \equiv&\neg\exists_x(P(x)) \land Q(y)\\ \equiv& \neg\exists_x(P(x)) \land Q(y)\tag{2} \end{align}
The boxed part is where you got confused, variables with same name in different Scopes are not guaranteed to be the same. In this case, second $x$ is not in any Scope, certainly not same as the first $x$.