Discussion on even and odd perfect numbers.

$1.$ By a theorem of Euler, any even perfect number is of the shape $2^{p-1}(2^p-1)$ where $p$ is prime. We don't need the primality part. Multiply by $8$, add $1$. We get $2^{2p+2}-2(2^{p+1})+1$, which is the square of $2^{p+1}-1$.

$2.$ We show that an odd perfect number cannot have only $2$ distinct prime factors (we don't deal with only $1$ prime factor, it's easier).

Let $N=p^aq^b$ where $p$ and $q$ are odd primes and $p\lt q$. Then the sum of the divisors of $N$ is $$(1+p+\cdots+p^a)(1+q+\cdots+q^b).$$ Using the ordinary formula for the sum of a finite geometric series, this can be rewritten as $$\frac{p^{a+1}-1}{p-1}\cdot \frac{q^{b+1}-1}{q-1}.$$ Divide by $N$. The result is $$\frac{p-\frac{1}{p^a}} {p-1}\cdot \frac{q-\frac{1}{q^b}} {q-1} .$$ This is less than $$\frac{p}{p-1}\cdot \frac{q}{q-1}.\tag{$1$}$$ We show that the product $(1)$ must be less than $2$. In particular, that shows the product cannot be $2$, so $N$ is in fact deficient.

It is easier to show that the reciprocal of Expression $(1)$ is greater than $\frac{1}{2}$. This reciprocal is $$\left(1-\frac{1}{p}\right)\cdot \left(1-\frac{1}{q}\right).$$

But $p\ge 3$ and $q\ge 5$. So $1-\frac{1}{3}\ge \frac{2}{3}$ and $1-\frac{1}{q}\ge \frac{4}{5}$. So their product is $\ge \frac{8}{15}$, which is greater than $\frac{1}{2}$.

I have not thought about your Question $3$.

As Andre mentioned, the even perfect numbers have the form $2^{p-1} (2^{p} -1 )$.

Claim: The sum of the first $2^{(p-1)/2}$ odd cubes is equal to this perfect number.

Proof: We know that the sum of the first $n$ odd cubes is:

$\sum_{i=1}^{n} (2i-1)^3 = \sum_{i=1}^{2n} i^3 - \sum_{i=1}^{n} (2i)^3 \\= \left[ \frac {(2n)(2n+1)}{2} \right] ^2 - 8\left[ \frac {n(n+1)}{2} \right]^2 = \frac {n^2 [16n^2 + 16n + 4 - 8(n^2 + 2n+1)]}{4}\\ = n^2(2n^2 - 1) $

Now substitute $n = 2^{(p-1)/2}$.