I am preparing for an exam on aspects of Logic related to propositional and first order logic. One of my revision exam questions is 
. I have attempted this question but I am really struggling with all parts of the question. If anyone could provide any hints or suggestions then I would be very grateful. My current (attempt) at a solution is:
Please note that for (a) I am suspicious of my proof because -> elimination is not used anywhere, for (b) my work hasn't progressed much beyond writing the definitions in the proof, and (c), I am struggling with the valuation extension.
$$ \{\alpha,\beta\} \vdash \alpha \qquad \mathrm{given} \\ \{\alpha\} \vdash \beta \to \alpha \qquad \mathrm{insertion} \\ \emptyset \vdash \alpha \to (\beta \to \alpha) \qquad \mathrm{insertion}$$
and since $\emptyset\subseteq\Sigma$ for all $\Sigma$, and inclusion preserves derivability, $\Sigma \vdash \alpha \to (\beta \to \alpha)$ for all $\Sigma$
For (b) just do, if $\Sigma$ is maximally consistent and $\Sigma\not\vdash\alpha$, then $$ \Sigma \cup \{\alpha\} \vdash \alpha\to(\alpha\to\beta) $$ and so, by elimination you get $\Sigma \vdash \alpha \to \beta$
Finally, for (c) you just start to define the valuation on your base cases $v(\phi)$ when $\phi\in X$, in which case you define $v(\phi) = \top $ if $\phi \in X\cap \Sigma$ and $v(\phi) = \bot$ if $\phi \in X\setminus \Sigma$ finally you define the recursion for formulae of the form $\phi\to\psi$ where you define $v(\phi\to\psi)=\top$ if $\phi\to\psi\in\Sigma$ and $v(\phi\to\psi)=(\lnot v(\phi))\lor v(\psi)$ otherwise.
To check that this valuation obeys the necessary precepts, you just need to show that the recursive step preserves the valuation when you do introduction and elimination, since by definition it already does so for insertion.