What is the ∆H_vap when temperature of a given liquid is equal to the boiling point of the liquid?

What is the $$∆H_{vap}$$ when temperature of a given liquid is equal to the boiling point of the liquid?

For the other cases ie (i) T>BP, $$∆H_{vap}$$ > 0 (ii) T<BP, $$∆H_{vap}$$ < 0

So what will be the $$∆H_{vap}$$ when T (Temperature) is exactly equal to BP of the liquid?

Because we know that heat has to be an input to convert from liquid phase to gaseous phase so can the the $$∆H_{vap}$$ = 0 ? If so, how is it possible since we are inputting heat into the container.

$$\Delta H_\mathrm{vap} \gt 0$$ is valid regardless of the relation $$T$$ vs $$T_\mathrm{vap}$$, as it is an endothermic process, requiring energy to break intermolecular bonds. Its value drifts with T, depending on the relation of the specific/molar heat capacity of the liquid and the vapour, following the Hess's Law.
You probably confuse $$\Delta H_\mathrm{vap}$$ which is always positive and $$\Delta G_\mathrm{vap} = \Delta H_\mathrm{vap} - T \cdot \Delta S_\mathrm{vap}$$, which is zero for the phase equilibrium for given $$T$$ or $$p$$ ( sitting on the (l)-(g) line of the phase diagram.)
If $$T \lt T_\mathrm{vap}$$ for the given $$p$$ then $$\Delta G_\mathrm{vap} \gt 0$$
If $$T \gt T_\mathrm{vap}$$ for the given $$p$$ then $$\Delta G_\mathrm{vap} \lt 0$$