First, let's solve the inequality for set $A$. We understand that the absolute value $|x-1|$ being greater than 2 means that the quantity inside the absolute value must be either greater than 2 or less than -2.

For $x-1 > 2$, we get:
\begin{align*}
x-1 &> 2 \\
x &> 3
\end{align*}

For $x-1 3 \text{ or } x < -1\}$.

Next, let's analyze set $B$. The inequality $x(x-5) < 0$ implies that the product of $x$ and $(x-5)$ is negative. This means that $x$ and $(x-5)$ have opposite signs. Based on the zero product property, the solution will be between the two roots of the quadratic equation, which are $x = 0$ and $x = 5$.

As a result, $B=\{x \mid 0 < x < 5\}$.

Now we need to determine the intersection $A \cap B$, which consists of elements that are common to both sets $A$ and $B$. The set $A$ includes numbers greater than 3 and less than -1, and the set $B$ includes numbers between 0 and 5.

Therefore, the intersection $A \cap B$ will only include numbers that are between 3 and 5, because that is the region where the conditions of both sets $A$ and $B$ are satisfied simultaneously.

The solution is:
$$A \cap B = \{x \mid 3 < x < 5\}$$

We can highlight the final answer as:
$$\boxed{ A \cap B = \{x \mid 3 < x < 5\} }$$