Let the first hyperbola be
$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, with eccentricity $e$.
And the second hyperbola be
$\dfrac{y^2}{B^2} - \dfrac{x^2}{A^2} = 1$ with eccentricity $E$.
Then the right focus of the first hyperbola is at $x = a e$
and the top focus of the second hyperbola is at $y = B E$
The y-coordinate of the upper end of the latus rectum of the first hyperbola is found by substituting $x = a e$ into the the equation of the hyperbola
$e^2 - \dfrac{y^2}{b^2} = 1$
so,
$y = b \sqrt{e^2 - 1}$
and this is equal to $BE$
i.e. $ B E = b \sqrt{e^2 - 1} $
and similarly
$ a e = A \sqrt{E^2 - 1}$
hence
$ B = \dfrac{b}{E} \sqrt{e^2 - 1}$
$ A = \dfrac{a e}{ \sqrt{E^2 - 1} }$
and we know that $E^2 = 1 + \dfrac{A^2}{B^2}$
so
$E^2 B^2 = A^2 + B^2 = \dfrac{b^2 (e^2 - 1)}{E^2} + \dfrac{a^2 e^2}{ (E^2 - 1)} $
But $E^2 B^2 = b^2 (e^2 - 1)$, therefore,
$b^2 (e^2 - 1) = \dfrac{b^2 (e^2 - 1)}{ E^2} + \dfrac{a^2 e^2}{(E^2 - 1)}$
Collecting terms,
$b^2 (e^2 - 1) ( 1 - \dfrac{1}{E^2}) = \dfrac{a^2 e^2}{(E^2 - 1)}$
now $e^2 = \dfrac{ (a^2 + b^2) }{ a^2 }$
hence
$\dfrac{b^2 (e^2 - 1) (E^2 - 1)}{ E^2 } = \dfrac{(a^2 + b^2)}{ (E^2 - 1)}$
Cross multiplying, and dividing through by $b^2$
$ \dfrac{(E^2 - 1)^2}{E^2} = \dfrac{(a^2 + b^2)}{( b^2 (e^2 - 1) ) }= \dfrac{( (a/b)^2 + 1 ) }{ (e^2 - 1) } $
but $\dfrac{b^2}{a^2} = e^2 - 1$, so $\dfrac{a^2}{b^2} = \dfrac{1}{(e^2 - 1)}$ , and $\dfrac{a^2}{b^2} + 1 = \dfrac{e^2 }{(e^2 - 1)}$
Hence,
$\dfrac{(E^2 - 1)^2}{E^2} = \dfrac{e^2}{(e^2 - 1)^2} $
Taking the square root,
$\dfrac{(E^2 - 1)}{E} = \dfrac{e}{(e^2 - 1)}$
thus
$(E - \dfrac{1}{E})(e - \dfrac{1}{e}) = 1 $
This proves that #2 is true.
Let's check if #4 is true or not. If $E = e$ then
$(e - \dfrac{1}{e})^2 = 1$
so
$e^2 + \dfrac{1}{e^2} - 2 = 1$
i.e. $e^2 + \dfrac{1}{e^2 }= 3$
or $(e^2)^2 - 3 (e^2) + 1 = 0$
and this gives
$e^2 = \dfrac{1}{2} ( 3 \pm \sqrt{5} )$
since $e \gt 1$ , then $e^2 = \dfrac{1}{2} (3 + \sqrt{5}) \approx 2.618 $
Therefore #4 is False.
Finally, we want to check the range for $eE$. We now know that
$(E^2 - 1)(e^2 - 1) = E e$
Hence,
$(E e)^2 - e^2 - E^2 + 1 = E e
define $u = E e$, and $v = \dfrac{E}{ e} $
then $ u v = E^2 $ and $ \dfrac{u}{v} = e^2 $
Hence,
$u^2 = u v + \dfrac{u}{ v} + u - 1 = u ( 1 + v + \dfrac{1}{v}) - 1 \gt 3 u - 1 $
Therefore, we now have,
$ u^2 - 3 u + 1 \gt 0 $ where $u > 1$
The roots of $u^2 - 3u + 1 = 0$ are $u \approx 0.3819$ and $ u \approx 2.618 $. Since u satisfies the above inequality , and $u \gt 1$ then $u \gt 2.618 $
hence $u \gt 2$, however $u$ is not necessarily greater than $3$. So #1 is true,but #3 is False.
An example of the given situation is depicted below with $e = 1.5$ and $a = 1 $