How to find the Cp, difference of Cp and heat of fusion from the dependence of vapor pressure on temperature? The vapor pressure of zinc varies with temperature as
$$\log P(\pu{mm Hg}) = -6850/T - 0.755\log T + 11.24 \label{eqn:1}\tag{1}$$
and that of liquid zinc as
$$\log P(\pu{mm Hg}) = -6620/T - 1.255\log T + 12.34 \label{eqn:2}\tag{2}$$
Calculate:
a) boiling point of $\ce{Zn};$
b) triple point;
c) heat of evaporation at boiling point;
d) heat of fusion;
e) the difference in $C_p$ of solid and liquid zinc.

I have found out the boiling point by substituting pressure as $\pu{760 mm Hg}$ on first equation \eqref{eqn:1}, from here boiling point $T$ can be found out as $\pu{1181 K}.$
I have also found out triple point by equating the two equations because at triple point both the pressure of the two phases must be equal.
I have also found out heat of vaporization by using the equation
$$\frac{\mathrm d\ln p}{\mathrm dT} = \frac{H_\mathrm{vap}}{RT^2}.$$
But to find the heat of fusion I must know the heat of sublimation, but the heat of sublimation is not provided here. Is there any other way to know the heat of fusion (melting)?
Also, how to find the $C_p$ of both the phases and the difference of $C_p?$
Use the equation for the vapor pressure of the solid to get the heat of sublimation.  The heat capacity of the vapor minus the heat capacity of the liquid is equal to the derivative of the heat of vaporization with respect to temperature. The heat capacity of the vapor minus the heat capacity of the solid is equal to the derivative of the heat of sublimation with respect to temperature.

ADDENDUM

$$\Delta H_{sublimation}=RT^2\frac{d\ln{P}}{dT}=(\ln{10})RT^2\frac{d\log{P}}{dT}$$where log P is given by Eqn. 1.