A light Ray falls on a square glass slab On horizontal face with incident angle 45. What must be the index of refraction if total internal reflection occurs at the vertical face.

Let the refractive index be RI
angle of from air to glass slab incidence = 45 degrees.
let R1 be the angle of refraction
sin(45 deg)/sin(R1) = Ri
sin(R1) = sin(45 deg)/(Ri)
angle of incidence of the ray on the vertical face of the slab as it travels through it is (90 degrees - R1)
sin(90)/sin(90 deg -R1) = Ri (here angle of refraction should be 90 degrees because we need total internal reflection)
1/cos(R1) = Ri
cos(R1) = 1/Ri

using sin square theta + cos Square theta = 1, we have square(sin(45 deg)/Ri) + square(1/Ri) = 1
square(1/(sqrt(2)*Ri)) + square(1/Ri) = 1
1/(2*square(Ri) + 1/square(Ri) = 1
=3/(2*square(Ri) ) = 1
square(Ri) = 3/2 Hence Ri = sqrt(3/2)

for second question, is the light ray travelling from air to water, or traveling from water to air? when you said 60 degrees, is it with respect to the surface or with respect to the normal? wording of the question is not very clear....

if i assume that the light ray is traveling from water to air, incident at an angle of 60 degrees with respect to normal, total internal reflection will occurs, because critical angle is sine inverse of 1/sqrt(2) = 45 degrees
angle of incidene = angle of reflection. so the reflected ray is 60 degrees to the normal, that is 30 degrees to the surface. if you extrapolate the incident ray in a straight line, it would make 30 degrees angle with the surface. so deviation would be 30 degrees + 30 degrees , = 60 degrees. if youy dray the diagram on the paper you would understand the answers better.