Nitric acid isn't strong enough, but ozone is, isn't it? Still, it doesn't oxidises it to periodic acid... can you explain why?

Can this mechanism be concerted? That is, the abstraction of proton, migration of phenyl and elimination of Br all in a single step?

Is the answer given, yes? I too think you are right, and also, tollen's reagent oxidises only compounds with free CHO group, like aldehydes and alpha hydroxy ketones, so why should this give positive tollen's test?

Why not just use one equivalent of $\ce{KMnO_4}$?

An $\ce{sp}$ hybridized species, which has a bond angle of $180^{\circ}$ is more electronegative than an $\ce{sp^2}$ hybridized species, which has a bond angle of $120^{\circ}$ followed by $\ce{sp^3}$ which has a bond angle of $109.5^{\circ}$. So, as bond angle increases, electronegativity increases. So, as you said that 1 has a greater bond angle than 2, it should be more electronegative and hence it would show greater inductive effect on $\overset{+}{\ce{N}}$.

@BuckThorn I just googled it and this is the first result I got. Of course I knew that it was not reliable. I couldn't find any good source. Also this link shows an experimental value of 8.82 for benzylamine.

I sincerely thank you for your efforts into this answer. The values that I gave were just based upon a google search and so I knew that these are not reliable values :).

@user55119 Sir please see my above comment, is my explanation for stable existence of free radical at bridge head positions correct? If not, why can a free radical exist at bridge head position then?

@SafdarFaisal: So is the answer to the question "why can free radicals form at bridge head carbon whereas carbocations cannot...?" is because a carbocation is $\ce{sp^2}$ hybridized whereas a free radical would not be $\ce{sp^2}$, it's not $\ce{sp^3}$ as well, but something in between. And $\ce{sp^2}$ hybridization is not allowed as per Bredt's rule. Am I right?

@BuckThorn I just did. So benzylamine has a pKa of 8.82 and benzyl alcohol has a pKa of 15.4 . Even in this case, I don't know the reason behind this. I would appreciate if answerer explains this case too. But why should this be related to the original question?

@Srini First you have written $-P\Delta V=-150$ which is fine. But when you did the calculation for $\Delta U$, you are subtracting -150, where 150 had to be subtracted.

You have made a mistake in sign of work as $P\Delta V=150$. Also, first law of thermodynamics is valid everywhere, not just at constant pressure.

@NisargBhavsar But we can come up with an explaination for the observation. So, that's what I am looking for. What's wrong with that?