Can any one of the following compound propositions can be created by using only ‘$\to$’ and ‘$\vee$’? 
Can any one of the following compound propositions can be created by using only ‘$\to$’ and ‘$\vee$’?
(i)  $ ¬ P $
(ii) $ P \wedge Q$
(iii) $ P \leftrightarrow Q$

My answer to (i)
Since  $¬$ is not a logical connective, it can not join two statements together.
Thus $¬$ can not be created by using $\to$ and $\wedge$.
Am I right? Also how to say true or false for the other two options?
Any help?
 A: You can show by induction on the complexity of the formula $\phi \equiv \phi_1 \vee \phi_2$ or $\psi \equiv \psi_1 \rightarrow \psi_2$ that the formula does not imply $\neg P$.
$\bf Base\ case.$ No atomic formula $Q$ implies $\neg P$.
$\bf Induction\ step\ 1.$ If either $\phi_1$ or $\phi_2$ fails to imply another formula $\theta$ (such as $\neg P$) then $\phi_1 \vee \phi_2$ also fails to imply $\theta$. (This may be easier to see from the counterfactual: if $\phi_1 \vee \phi_2$ implies $\theta$, then both $\phi_1$ and $\phi_2$ do as well.)
$\bf Induction\ step\ 2.$ If $\psi_2$ fails to imply $\theta$ then $\psi_1 \rightarrow \psi_2$ fails to imply $\theta$. (If you like, this can be because $\psi_1 \rightarrow \psi_2$ is equivalent to $\neg \psi_1 \vee \psi_2 $, which lets us reduce the $\rightarrow$ case to the $\vee$ case.)

Question for you: does this strategy also work for (ii) $P \wedge Q$ and (iii) $P \leftrightarrow Q$? Or can you solve these cases by some other means, e.g. the sort of argument  John Griffin has used in the comments to show (ii) implies (iii)?

Your reasoning that binary connectives be can't say anything about unary formulas is not correct in general. A famous and popular example of this behaviour is that $(P \rightarrow Q) \rightarrow P$ actually implies $P$.
