(1) is certainly not true for general signed measures $\mu$. However, if we restrict to signed measures with $\mu(X)=0$, then it is true with a factor of $2$, i.e. $$\|\mu\|_{TV} = 2 \sup_{A \in \mathcal{B}} |\mu(A)| \tag{*}.$$ That is, in this special case, the leftmost inequality in (2) is attained.
For one inequality, let $X = B^+ \cup B^-$ be the Hahn decomposition for $\mu$. Note that $\|\mu\|_{TV} = |\mu(B^+)| + |\mu(B^-)|$, while $\mu(X) = \mu(B^+) + \mu(B^-) = 0$ so that $\mu(B^+) = -\mu(B^-) = \frac{1}{2} \|\mu\|_{TV}$. Hence taking $A = B^+$ shows the $\le$ inequality in (*).
Conversely, for any $A \in \mathcal{B}$ we have $$\mu(A) = \mu(A \cap B^+) + \mu(A \cap B^-) \le \mu(A \cap B^+) \le \mu(B^+) \le \frac{1}{2} \|\mu\|_{TV}$$ and similarly $\mu(A) \ge -\frac{1}{2} \|\mu\|_{TV}$, so that $|\mu(A)| \le \frac{1}{2} \|\mu\|_{TV}$. This shows the $\ge$ inequality.
In particular, taking $\mu = P-Q$ where $P,Q$ are both probability measures, we see that $d_{TV}(P,Q)$ as defined by (3) is exactly half of $\|P-Q\|_{TV}$. So the definitions are the same, up to a constant factor of 2.