1993 AP Exam problem #6.
a) what angle does the reflected ray make the with normal to the surface?
same angle as the incident ray:
theta_water = arctan(2/3) = 33.69 deg

b) what angle does the emerging ray make with the normal to the surface?
Snell's Law:
sin(theta_water) * n_water = sin(theta_air) * n_air
sin(theta_air) = sin(theta_water) * n_water / n_air
theta_air = arcsin( sin(theta_water) * n_water / n_air )
    = arcsin( sin(33.69) * 1.33 / 1 )
    = 47.54 degrees

c) what is the minimum depth of water for which the light that strikes the surface of the water 2.0 meters to the left of the light source will be refracted into the air?

this refers to total internal reflection, so theta_air = 90 degrees.
sin(theta_water) * n_water = sin(theta_air) * n_air
sin(theta_water) * n_water = 1 * n_air
theta_water = arcsin(n_air / n_water)
   = arcsin(1/1.33)
   = 48.75 degrees
now to solve a triangle for depth given base of 2 m.
tan(48.75) = 2 / d
d = 2 / tan(48.75) = 1.754 m