What are the changes in free energy, enthalpy and entropy and the sign of work for the following process carried out at 100.8o C. The boiling point of HCOOH is 100.8o C. (-, +, or 0)

HCOOH (g) → HCOOH (l)

I knew that change in enthalpy and change in entropy is both zero because the calculation between T final and T initial from their equations make it to zero. But what about change in free energy and work? Please explain to me because I don't quite get how to decide if it has positive or negative value!

+ve value denotes an endothermic reaction, where the substance absorbs energy.

-ve value denotes an exothermic reaction, where the substance releases energy.

Work can be quite difficult to deduce the way you posted the question...
work done ON the substance is +ve.
work done BY the substance is -ve.

I'm not used to free energy, but from what I understand, is the energy that you can extract from the substance.

One simple fact to think about. You cooled the gaseous methanoic acid to liquid. Since you cooled it, it should have less energy, and hence, the free energy will be less. Hence I would say free energy change is -ve.