The enthalpy of sublimation is the total amount of energy required to convert a solid directly into a gas. This can be calculated by summing the enthalpy of fusion (solid to liquid) and the enthalpy of vaporization (liquid to gas). Mathematically, this relationship is represented as:

$$ \Delta H_{\text{sublimation}} = \Delta H_{\text{fusion}} + \Delta H_{\text{vaporization}} $$

Given:

Enthalpy of fusion, $$\Delta H_{\text{fusion}} = 2.8 \, \text{kJ mol}^{-1}$$

Enthalpy of vaporization, $$\Delta H_{\text{vaporization}} = 98.2 \, \text{kJ mol}^{-1}$$

Substitute these values into the equation:

$$ \Delta H_{\text{sublimation}} = 2.8 \, \text{kJ mol}^{-1} + 98.2 \, \text{kJ mol}^{-1} = 101.0 \, \text{kJ mol}^{-1} $$

Therefore, the enthalpy of sublimation of the substance (X) is approximately 101 kJ mol^$$-1$$.