1. I think you may have meant to say : \(\displaystyle \sum_{d|n}{phi(n)} = n\) in which case you can work it out like so - a method by Gauss that I learned from my book - and I'll go through it step by step since I am new to this myself:

Suppose \(\displaystyle n = 12\) and we have a class of numbers \(\displaystyle C_{d}\) where \(\displaystyle C_{d}\) holds any number less than \(\displaystyle n\) that has its highest divisor as \(\displaystyle d\). Then we have a set of classes:

\(\displaystyle C_{1} = \{1, 5, 7, 11\}\)

\(\displaystyle C_{2} = \{2, 10\} \)

\(\displaystyle C_{3} = \{3, 9\} \)

\(\displaystyle C_{4} = \{4, 8\} \)

\(\displaystyle C_{6} = \{6\} \)

\(\displaystyle C_{12} = \{12\}\)

A number m only exists in \(\displaystyle C_{d}\) if \(\displaystyle (m, n) = d\). Further, \(\displaystyle (m/d, n/d) = 1\) so a number m only exists in \(\displaystyle C_{d}\) if \(\displaystyle m/d\) and \(\displaystyle n/d\) are relatively prime. By definition the number of integers less than \(\displaystyle n/d\) and relatively prime to \(\displaystyle n/d\) is \(\displaystyle \phi(n/d)\) so the number of elements in \(\displaystyle C_{d}\) is \(\displaystyle \phi(n/d)\).

From here it is a simple matter.

We just proved that \(\displaystyle \sum_{d|n}{phi(d/n)} = n\) and if you work it out manually with a small number like \(\displaystyle n = 12\) you will find that \(\displaystyle \sum_{d|n}{phi(d/n)} = \sum_{d|n}{phi(n)}\) and with that we have proved \(\displaystyle \sum_{d|n}{phi(n)} = n\)

For the record I relied quite heavily on my book for much of the above so I cannot take credit for this work.

2. I know there is a much more complete way of proving this, but at the moment I can only think of a proof by contradiction:

Lets suppose \(\displaystyle \phi(2n) = \phi(n)\) works for even numbers as well. If \(\displaystyle n = 2^1\), then we can use the equation \(\displaystyle \phi(p^{e}) = p^{e-1}(p-1)\) where p is a prime and e is its power. If we work it out with \(\displaystyle n = 2\) (a prime) then \(\displaystyle \phi(4) = 2^0(2-1) = 2\) while \(\displaystyle \phi(2) = 1\). We have reached a contradiction and we at least know for the even number 2 this does not work. The only problem is I do not know how to extend this to all even numbers. This is a very weak proof and I would like to see someone else write one better.

3. I can't figure this one out either.