How do you solve and write the following in interval notation: #1/|x+5| >2#?

First of all, we have to exclude the value when the inequality is undefined because the denominator is zero:
#x != -5#

In all cases, except this, #|x+5| > 0#.

Both sides of an inequality can be multiplied by a POSITIVE number, leaving the sign of inequality as is.

Since #|x+5| > 0# (after we have excluded value #x=-5#), let's multiply left and right sides of our inequality by #|x+5|# getting
#(1 * |x+5|) / (|x+5|) > 2 * |x+5|#
or
#1 >2*|x+5|#

Next transformation is division of both sides of inequality by POSITIVE number #2# preserving the sign of inequality:
#1/2 > |x+5|#

Recall the definition of #|a|#:
#|a| = a# for #a >= 0# and
#|a| = -a# for #a < 0#

Consider now two cases (we excluded #x=-5#):

Case 1. #x+5 > 0# or, equivalently, adding #-5# to both parts of inequality preserving the sign of inequality, #x > -5#
Then #|x+5| = x+5# and our inequality looks like
#1/2 > x+5#
or, subtracting #5# from both sides and retaining the sign of inequality, #x < -9/2#
We have to combine this with an inequality that defines our case,
#x > -5#.
Both inequalities result in'
#-5 < x < -9/2#

Case 2. #x+5 < 0# or, equivalently, adding #-5# to both parts of inequality preserving the sign of inequality, #x < -5#
Then #|x+5| = -x-5# and our inequality looks like
#1/2 >- x-5#
or, add #x# to both sides and retaining the sign of inequality, #x +1/2 > -5#
or, subtracting #1/2# from both sides,
#x > -11/2#
We have to combine this with an inequality that defines our case,
#x < -5#.
Both inequalities result in'
#-11/2 < x < -5#

Final solution is
#-11/2 < x < -5# and #-5 < x < =9/2#

Here is an illustrative graph of a function #y= 1/|x+5|-2#. Notice, where it is positive.
graph{1/|x+5|-2 [-10, 2, -5, 5]}