Find cos(A+B) given cosA=1/3 , sinB=1/4, Where 0 < Pi/2 and Pi/2 < B < Pi

I assume you mean 0 < A < π/2

sin²A + cos²A = 1
sin²A + 1/9 = 1
sin²A = 8/9
sinA = √8/3 = 2√2/3

sin²B + cos²B = 1
1/16 + cos²B = 1
cos²B = 15/16
cosB = -√15/4
(take the negative root because B is in the second quadrant)

cos(A+B) = cosAcosB - sinAsinB
= 1/3 x -√15/4 - 2√2/3 x 1/4
= -√15/12 - 2√2/12
= -(√15 + 2√2)/12

so the equation is as follows:
cos(A+B) = (cosA x cos B) - (sinA x sinB)
theyve given you cos A and cos B so plug those in, but what about cos B and sin A:
you can use the Pythagorean theorem which is a-squared plus b-squared = c-squared. you already have a and c, just solve for b using values from sinB and cosA respectively.

SO:
cos(A+B) = (cosA x cos B) - (sinA x sinB)
= (1/3 x -sqrt(15)/4) - (sqrt(8)/3 x 1/4)
^yep thats negative squareroot 15 since the restriction on B says its in the upper left quadrant

**not entirely sure why you're telling us 0 is less than pi/2 but okay....was that supposed to be a restriction on sin?

cos (a + b) = cos(a)cos(b) - sin(a) sin(b)
= (1/3) cos(b) - (1/4)sin(a)

if sin(b) = 1/4 then cos^2(b) = 1 - sin^2(b) = 1- 1/16 =15/16
cos(b) = -sqrt(15)/4 (since it's in the second quadrant

if cos (a) = 1/3 then sin^2(a) = 1 - 1/9 = 8/9
sin(a) = 2sqrt(2)/3 since a is in the first quadrant

substitute in the cos(a+b) equation and you have the answer.

cos(A+B)=cosA*cosB-sinA*sinB
sinA=√￣（1-(1/3)^2）=2√2/3
cosB=-(√15)/4

so cos(A+B)=-(√15)/12-(√2)/6