showing $\neg\alpha\vee\delta,\neg\beta\vee\neg\delta\vdash \neg(\alpha\vee\beta)\vee\delta$ is valid Given tertium non datur ($\neg\alpha\vee\alpha$) and:
\begin{align} 
\beta&\vdash\alpha\vee\beta\tag{1}\\
\alpha\vee\alpha&\vdash\alpha\tag{2}\\
\alpha\vee(\beta\vee\delta)&\vdash(\alpha\vee\beta)\vee\delta\tag{3}\\
\alpha\vee\beta,\neg\alpha\vee\delta&\vdash\beta\vee\delta\tag{4}
 \end{align}
I want to show:
(i) $\alpha\vee\beta\vdash\beta\vee\alpha$;
 (ii) $\alpha\vee\beta\vdash\neg\neg\alpha\vee\beta$; and
 (iii) $\neg\alpha\vee\delta,\neg\beta\vee\neg\delta\vdash \neg(\alpha\vee\beta)\vee\delta$.
The first one is easy. Given $\alpha\vee\beta$ and $\neg\alpha\vee\alpha$ I get with (4) $\beta\vee\alpha$.
But I have no idea for (ii) and (iii), they seem to be quite hard. How can you show it? 
 A: Ad (ii): Use $\neg\alpha\lor \neg\neg\alpha$ and $(4)$.
Ad (iii): It is not strange that you have difficulty proving this: It is not true. Suppose $\beta,\neg\alpha,\neg\delta$. Then its premises are true, but its conclusion is false. 
But let us suppose that you meant for the second premise to be $\neg \beta\lor\delta$.
Then we can do the following (using (i) and $(3)$ repeatedly without mention):
$$\begin{align}
\neg\alpha \lor \delta, \neg(\alpha\lor\beta)\lor(\alpha\lor\beta)&\vdash \beta\lor\neg(\alpha\lor\beta)\lor\delta\\
\neg\beta\lor\delta,\beta\lor\neg(\alpha\lor\beta)\lor\delta&\vdash \neg(\alpha\lor\beta)\lor\delta\lor\delta
\end{align}$$
To conclude, we need to prove:
$$\neg(\alpha\lor\beta)\lor\delta\lor\delta\vdash\neg(\alpha\lor\beta)\lor\delta$$
Fortunately, this isn't very hard (once you see it, I might add):
\begin{align}
\neg(\alpha \lor \beta)\lor \delta\lor\delta &\vdash \neg(\alpha\lor\beta)\lor\delta\lor\delta\lor\neg(\alpha\lor\beta)\\
&\vdash (\neg(\alpha\lor\beta)\lor\delta)\lor(\neg(\alpha\lor\beta)\lor\delta)\\
&\vdash \neg(\alpha\lor\beta)\lor \delta
\end{align}

In fact, using weakening ($\beta\vdash\alpha\lor\beta$), contraction ($\alpha\lor\alpha\vdash\alpha$), association ($(3)$) and commutation ((i)) we can show that: $$\bigvee_{i=1}^m \alpha_i \vdash \bigvee_{j=1}^n \beta_j$$ as soon as: $$\{\alpha_i: 1\le i\le m\} \subseteq \{\beta_j: 1 \le j\le n\}$$
Together with (ii) and (iii), this result is used by Shoenfield to prove (by structural induction) the Tautology theorem (paraphrased):

If $\neg\alpha_1\lor\cdots\lor\neg\alpha_n \lor\beta$ is a (truth-tabular) tautology, then $\alpha_1,\ldots,\alpha_n\vdash\beta$.

An important intermediate step for this result is:

If $\alpha_1\lor\cdots\lor\alpha_n$ is a tautology, then $\vdash \alpha_1\lor\cdots\lor\alpha_n$.

and it is this "lemma" that is proved using the three ingredients mentioned above.
