1) tan(sin^-1(5/6)-cos^-1(1/7)

Question

2) tan(è + ); cos è = − 1/7,è in Quadrant III, sin = 1/2, in Quadrant II

Answer 1

Let A = sin^-1 (5/6) and B = cos^-1 (1/7)
then
tan(sin^-1(5/6)-cos^-1(1/7)
= tan(A-B)
= (tanA - tanB)/(1 + tanAtanB)

from A = sin^-1 (5/6)
sinA = 5/6, then cosA = √11/6 and tanA = 5/√11

from B = cos^-1 (1/7)
cosB = 1/7, then sinA = √48/7 and tanB = √48/1 = √48

then tan(A-B)
=(5/√11) - √48)/(1+(5/√11)(√48) )
= ( (5-√528)/√11) / (√11 + 5√48))/√11 )
= (5 - √528)/(√11 + 5√48)

= (5-4√33)/(11+20√3)

I cannot decipher your second question