#### Answer

$ a.\quad$Not necessarily true.
$ b.\quad$Necessarily true.
$ c.\quad$Necessarily true.
$ d.\quad$Necessarily true.
$ e.\quad$Necessarily true.
$ f.\quad$Necessarily true.
$ g.\quad$Necessarily true.
$ h.\quad$Necessarily true.

#### Work Step by Step

In interval notation, the set $\{x|2\lt x \lt 6\} $represents the open interval $ I=(2,6), \quad$ containing numbers between 2 and 6 (but without 2 and 6).
$ a.\quad$
This interval, $(0,4)$ contains numbers between $0$ and $4 $ (no borders).
There are numbers $x$ from interval $I$, such as $x=5$, for which this statement is not true.
And, there are numbers from $I$, such as $x=3$, for which this statement holds true.
Thus, it is not necessarily true.
$ b.\quad$
By rule 1 from the "Rules for Inequalities" table, we can add 2 to each part of the compound inequality:
$0+2 \lt x-2+2 \lt 4+2$
$2 \lt x \lt 6$,
which is true for any $x$ we choose from the interval $I.$
Necessarily true.
$ c.\quad$
By rule 3 from the "Rules for Inequalities" table,
$1 \displaystyle \lt \frac{x}{2} \lt 3,\qquad $
multiply all parts with $2$:
$2 \lt x \lt 6$,
which is true for any $x$ we choose from the interval $I.$
Necessarily true.
$ d.\quad$
By rule 6 from the "Rules for Inequalities" table,
$6\gt x \gt 2$
which is another way of writing
$2 \lt x \lt 6$,
which is true for any $x$ we choose from the interval $I.$
Necessarily true.
$ e.\quad$
If $x $is taken from the inteerval $I$, it is positive, so we can apply rule 6 from the "Rules for Inequalities" table,
$1\displaystyle \gt \frac{x}{6} \gt \frac{1}{3}\quad $
apply rule 3 (multiply with 6)
$6\gt x \gt 2$
which is another way of writing
$2 \lt x \lt 6$,
which is true for any $x$ we choose from the interval $I.$
Necessarily true.
$ f.\quad$
Applying property 5 from the table "Absolute values and intervals",
$-2\lt x-4 \lt 2 \quad $
apply rule 1 ( add 4 to each part)
$2 \lt x \lt 6$,
which is true for any $x$ we choose from the interval $I.$
Necessarily true.
$ g.\quad$
By rule 4, multiplying with $(-1)$ we obtain
$6\gt x \gt -2$
which is another way of writing
$-2 \lt x \lt 6$,
which is true for any $x$ we choose from the interval $I$, because
$-2\lt 2 \lt x \lt 6$
(The interval $I$ is a subset of this interval)
Necessarily true.
$ h.\quad$
By rule 4, multiplying with $(-1)$ we obtain
$6\gt x \gt 2$
which is another way of writing
$2 \lt x \lt 6$,
which is true for any $x$ we choose from the interval $I.$
Necessarily true.