Using truth table for validity Just need a quick clarification as to what this means, I think I understand the method to do it
$$(A \lor B), C \models_{\lower{1ex}{\rm taut}} (A \land B) \to (A \lor C)$$
Does this mean that $(A ∨ B)$ and $C$ are both assumed to be true, THEN I have to prove that $(A ∧ B) → (A ∨ C)$ is always true?
 A: Yes, this is what it means:
$\psi_1, ..., \psi_n \vDash \phi$
iff 
for all assignments $v$: If $[[\psi_i]]_v = 1$ for all $\psi_i, 1 \leq i \leq n$, then $[[\phi]]_v = 1$.
I.e., you only consdier the assignments under which all of the premises are true, and make sure that under these assignments, the conclusion holds as well. If, under a particular assginment, not all of the premises are true (= one or more of the premises is false), then the concolusion may be true or false.
The entailment does not hold iff there is at least one assignment under which all of the premises are true, but the conclusion is not.
Thus, in your truth table, you need to consider those rows where both $A \lor B$ and $C$ are true, and check whether $A \land B \to A \lor C$ holds as well for all of these rows. Rows where at least one of the premises is false may be disregarded.
Edit: Just to clarify, because your wording is a bit ambiguous: It doesn't mean that $\phi$ must be "always true" in the sense of a tautology: It must only be true under those assignments under which all of the premises are true.
This means that $\phi$ may even be unsatisfiable, in case that the conjunction of the premises is unsatisfiable (because then, there is no case under which the premises are true and the conclusion is not true, which would be needed as a counterexample to disprove the entailment relation).
