The entropy of the system decreases
All the reactants are mono-atomic gases, and the product is a solid. This is a strong indication that the entropy of the system decreases.
The reaction is spontaneous
That statement from the text of the question means that the Gibbs energy of reaction is negative. It is not clear if this refers to standard conditions, or to the current conditions. The question could be more specific in that regard.
Enthalpy of reaction
If the reaction entropy is negative and the reaction Gibbs energy is negative, the reaction enthalpy has to be negative as well because it is the sum of the Gibbs energy and the entropy times the (positive) temperature.
$$\Delta_r G = \Delta_r H - T \Delta_r S$$
$$\Delta_r H = \Delta_r G + T \Delta_r S$$
Are all exothermic reactions spontaneous?
No, but all reactions that are spontaneous at standard state (interpreted to mean that at standard state, equilibrium lies in the forward direction, i.e. K > 1 and $\Delta_r G^\circ < 0$) are either exothermic ($\Delta_r H^\circ < 0$) or show a positive standard entropy of reaction ($\Delta_r S^\circ > 0$), or both. This ensures that the Gibbs energy of reaction is negative and that the entropy of the universe increases (those two are linked in the absence of non-PV work).
[theorist] The direction of spontaneity is merely a function of where the reaction mixture is relative to the equilibrium constant.
The argument I made above does not preclude finding a reaction with positive $\Delta_r H^\circ$ and negative $\Delta_r S^\circ$ that goes forward. In this case, K will be smaller 1, but reaction conditions can be chosen so that Q is smaller than 1 as well, and furthermore smaller than K, which is always possible by increasing reactant concentrations or by removing product.
[Textbook question]Pick the answer that best describes this process.
Answers A) and E) are linked. An endothermic process (A) will decrease the temperature of the system (E). So those two answers are out if there is only one correct answer. Answer C) goes against the second law of thermodynamics, so most teachers would mark it wrong. Above, I made an argument that the reaction has a negative standard entropy of reaction (less freedom for the atoms when they go from mono-atomic gas to a solid compounds); answer D) posits the opposite, so it is out.