for question 3)

(x-(1-x²)^1/2)^1/2

first of all (1-x²)>=0

=> x   [-1,1] --------------------------------------------1

next (x-(1-x²)^1/2) >=0  

=> x >= 0 -------------------------------------------------2

&

x >= (1-x²)^1/2 

since terms on either side of the inequality are positive hence the trms can be squared hence

=> x² >= 1- x² 

=> 2x² >= 1

=> x² >= 1/2

=> x  [-infinity,-1/root(2)] U [1/root(2),infinity) -------------------------- 3

therefore by taking intersection of  1 , 2 & 3

domain is [1/root(2) , 1]