The reaction a(g)⇌b(g) has an equilibrium constant that is less than one. part a what can you conclude about δg∘rxn for the reaction?

The equilibrium constant of reaction, usually denoted as K, is a unit of ratio. The ratio involves concentrations or partial pressures of products to reactants. But you also have to incorporate their stoichiometric coefficients in the reaction as their respective exponents. If K is in terms of concentration, only the substances in their aqueous state are the ones that are included only in the expression. If K is in terms of partial pressures, only the substances in gaseous states are the ones that are included only in the expression. For this problem, it would be in terms of partial pressures. To properly show you how it's done, consider this equilibrium reaction:

aA (g) + bB (g) ⇆ nN (g)

The equilibrium constant for this reaction is:
K = [N]ⁿ/[A]ᵃ[B]ᵇ
where the [] brackets denotes partial pressures of the substances

Particularly, for the reaction a(g)⇌b(g), the K expression would be
K = [B]/[A]
So, if K is less than one, that means that the numerator is less than the denominator. It follows that the partial pressure of reactant A is greater than product B. Since A is greater, then the more favorable direction would be the forward reaction. The δG°rxn would then be negative in value. So δG°rxn < 0. 

To explain, δG°rxn is a criterion for spontaneity. If δG°rxn is negative, the reaction is spontaneous. If δG°rxn is positive, it is non-spontaneous. Since the favorable reaction is the forward reaction, it is spontaneous.

B. a graduated substance