How do you find the slope of a line that is a) parallel to the line given and b)perpendicular to the line with the given equation #2x-5y = -7#?

If a line is expressed by the equation #y=mx+c# then #m# is the slope and #c# is the intercept.

Starting with the equation given, first add #5y# to both sides to get:

#2x=5y-7#

Add #7# to both sides to get:

#2x+7 =5y#

Divide both sides by #5# to get:

#2/5x+7/5=y#

In other words #y = 2/5x + 7/5#

Comparing with #y=mx+c#, we see that the slope is #2/5# and the intercept is #7/5#.

Any line parallel to this line will have the same slope: #2/5#.

Any line perpendicular to a line of slope #2/5# will have a slope of #-5/2#.

There are several ways to see this.

For example, consider the following steps:
(1) Reflect the original line in the line #y = x#. This will have the effect of swapping #x# and #y# in the original equation.
(2) Reflect the line in the #x# axis, given by the equation #y = 0#. This will have the effect of inverting the sign of the #y# coordinate.

The total effect of (1) followed by (2) is a rotation of #pi/2#. That is, the resulting line is perpendicular to the line you started with.

Starting with your original line this will result in a line with equation #-2y-5x=-7#, which can be reformulated as:
#y = -(5/2)x+7/2#. This has slope #-5/2# as predicted.