Chemistry - Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation?

It very much depends on what definition of the equilibrium constant you are looking. The most common usage of the same has quite a variety of possible setups, see goldbook:

Equilibrium Constant
Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always unitless, as it is defined differently (goldbook)

Standard Equilibrium Constant $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant) Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{RT}\right\}$$ where $\Delta_rG^\circ$ is the standard reaction Gibbs energy, $R$ the gas constant and $T$ the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

It is worthwhile to note, that the first definition is always an approximation to the standard definition and in the standard definition all compounds regardless of their state are included in the equilibrium.

If you are looking at a reaction that takes place in gas phase, solid materials will play a constant role, since their partial pressure will solely depend on their vapour pressure and can therefore be regarded as constant. Therefore it can be seen as a part of the equilibrium constant. (The same applies to liquids while deriving for gas phase.) If you would increase the amount of solid in the system you would still not change it's concentration in gas phase.

For the reaction $$\ce{H2 (g) + I2 (s) <=> 2HI (g)}$$ you can form the standard equilibrium constant with activities/ fugacities $$K^\circ = \frac{a^2(\ce{HI})}{a(\ce{H2})\cdot{}a(\ce{I2})},$$ with $$a=\frac{f}{p^\circ}.$$ The activity for a pure solid is normally defined as one ($a(\ce{I2})=1$) and therefore $$K^\circ\approx K = \frac{a^2(\ce{HI})}{a(\ce{H2})}$$

For concentration dependent equilibrium constants, the following assumption is to be considered: $c(\ce{H2O})\approx55.6~\mathrm{mol/L}$ and is usually always much larger than that of any other component of the system in the range where the equilibrium approximation can be used. Also in most solution based reactions, the solute itself does not directly influence the reaction. Its concentration will therefore not change (significantly) and can be included in the equilibrium constant.

You are correct in assuming that the kinetics of a surface reaction will only depend of the actual area of the surface (and of course the reactants forming the products). But here again this area is most likely to be considered as constant and will result in a scalar for the reaction rate (in the area where an accurate description of the reaction is possible).

As of Greg's comment, He is of course right. In an equilibrium process the surface area will affect both directions of the reaction, hence it will not show up in the reaction rate at all. (The only effect it might have is, that the equilibrium state will be reached faster.)