Step 1: Identify the element type.
The formation of a binary cation \( (EH_x)^+ \) indicates that element \(E\) can form stable protonated species, which is characteristic of group 15 elements.
Step 2: Use the chemical test.
The formation of basic mercury(II) amido-iodide with \(K_2HgI_4\) in alkaline medium is a confirmatory test for ammonia and ammonia-like hydrides, again indicating a group 15 element.
Step 3: Match ionisation enthalpy.
The first ionisation enthalpy values of the first elements of groups are approximately:
\[
\text{Group 13: } 801,\quad
\text{Group 14: } 1086,\quad
\text{Group 15: } 1402,\quad
\text{Group 16: } 1312.
\]
Thus, element \(E\) belongs to group 15, and its first ionisation enthalpy is
\[
1402\ \text{kJ mol}^{-1}.
\]
Final Answer:
\[
\boxed{1402}
\]