(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 $A \in \mathcal{B}$, the defining property of the Hahn decomposition implies $\mu(A \cap B^-) \le 0$ and $\mu(A^c \cap B^+) \ge 0$, and therefore we have $$\mu(A) = \mu(A \cap B^+) + \mu(A \cap B^-) \le \mu(A \cap B^+) \le \mu(B^+) = \frac{1}{2} \|\mu\|_{TV}.$$ A similar argument shows $\mu(A) \ge -\frac{1}{2} \|\mu\|_{TV}$, 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.
Your equation (4) is also off by a factor of 1/2. The identity $$\|\mu\|_{TV} = \sup_{\|f\|_\infty \le 1} \int f\,d\mu$$ is true for every signed measure. To see one direction, write $$\int f\,d\mu = \int f\,d\mu^+ - \int f\,d\mu^- \le \mu^+(X) + \mu^-(X) = \|\mu\|_{TV}.$$ For the opposite inequality, take $f = 1_{B^+} - 1_{B^-}$.