Why do we need three equations to find the pH of NaCN, given Ka(HCN)?
So strictly speaking you only really need equation 3 to solve this problem, but the first two equations help you figure out how to solve it if you're not too familiar. What do I mean:


  Equation 1: $\ce{HCN + H2O &lt;=&gt; H3O+ + CN-}$


As you correctly note, this is to remind you of the definition for $K_a$


  Equation 2: $\ce{NaCN -&gt; Na+ + CN-}$


Note that I've changed $\ce{&lt;=&gt;}$ to $\ce{-&gt;}$. This is because most sodium salts fully dissociate in aqueous solution and so this "equilibrium" doesn't exist and you can then consider all $\ce{NaCN}$ to be converted to $\ce{CN-}$ for your "initial" stage in the ICE table. Hence there's no need to write an ICE Table with respect to Equation 2.

Value of writing this out: 
In the case that you were given something sparingly soluble or did not fully dissociate, you would have to consider another equilibrium and this equation 2 would become significant. But in this case, $\ce{Na+}$ is a spectator ion which you can ignore.


  Where does Equation 3 come from?


Equation 3 comes from the question itself (or rather what is happening in the system described). Remember, you've got a 1.0M solution of $\ce{NaCN}$. Since we've already established that it fully dissociates, at the "initial" stage of the system you have only $\ce{Na+}$ and $\ce{CN-}$ ions floating around. But from Equation 1 you know that $\ce{HCN}$ is a weak acid and so free $\ce{CN-}$ in water may act as a base and take a proton from water, and that is given by the equation:

$$\ce{CN- + H2O &lt;=&gt; HCN + OH-}$$