Question 249059
Please remember one thing and your life will be easier:
*** Dividing two fractions is the same as multiplying the first fraction by the inverse of the second one. ***


Let's get to work:
1) {{{((-2)/(ab^2))/(5/a)}}}
The reciprocal of {{{5/a}}} is {{{a/5}}}:
{{{((-2)/(ab^2))*(a/5)}}}
Multiply numerators and denominators:
{{{(-2a)/(5ab^2)}}}
Now cancel variable "a":
{{{(-2)/(5b^2)}}}  <--- Here's the solution for problem #1


2) {{{((3y+2)/(7y^3))/((y-1)/(y^2))}}}
{{{((3y+2)/(7y^3))*((y^2)/(y-1))}}}
{{{((3y+2)/(7y*y^2))*((y^2)/(y-1))}}}
{{{((3y+2)/(7y))*((1)/(y-1))}}}
{{{(3y+2)/(7y^2-7y)}}}  <--- Here's the solution for problem #2


The next one is a little different.  We need to find the common denominators in order to add the fractions (in the numerator as well as in the denominator), then we will worry about multiplying by the reciprocal.
3) {{{((1/(3y))+(1/(6y)))/((1/(2y))+(3/(4y)))}}}
{{{((1/(3y))+(1/(6y)))/((1/(2y))+(3/(4y)))}}}
{{{(((1/(3y))(6y/(6y)))+((1/(6y))(3y/(3y))))/(((1/(2y))(4y/(4y)))+((3/(4y))(2y/(2y))))}}}
{{{((6y/(18y))+(3y/(18y)))/((4y/(8y))+(6y/(8y)))}}}
{{{(9y/(18y))/(10y/(8y))}}}
Multiply by reciprocal:
{{{(9y/(18y))*(8y/(10y))}}}
Reduce fractions:
{{{(1/(2y))*(4y/(5y))}}}
Multiply numerator and denominator
{{{(4y)/(2y*(5y))}}} ...same as... {{{(2y*2y)/(2y*(5y))}}}
Cancel 2y (num & den)
{{{2y/(5y)}}}
Cancel y
{{{2/5}}}


4) I could not understand how the fourth problem was supposed to be setup.  I'm having a hard time undertanding the denominator part of it.  Please try solving it by following previous problems.  However, if you get lost, send me an email using the "thank you" feature.