Answer:
1) iii i= 1m, 2)  iii and iv, 3)  i = f₂ (L-f₁) / (L - (f₁ + f₂))
Explanation:
Problem 1
For this problem we use two equations the equations of the focal distance in mirrors
               f = r / 2
               f = 2/2
              f = 1 m
 
The builder's equation
            1 / f = 1 / o + 1 / i
Where f is the focal length, "o and i" are the distance to the object and the image respectively.
For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0
               1 / f = 0 + 1 / i
               i = f
               i = 1 m
The image is formed at the focal point
The correct answer is iii
Problem 2
For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat
Case 1 Flat lens - convex (convergent)
               R₂ = infinity
               R₁ > 0
Cas2 Flat-concave (divergent) lens
              R₂ = infinity
               R₁ <0
Why the correct answers are iii and iv
Problem 3
For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface
           1 / f₂ = 1 / (L-f₁) + 1 / i
           1 / i = 1 / f₂ - 1 / (L-f₁)
            1 / i = (L-f₁-f₂) / f₂ (L-f₁)
            i = f₂ (L-f₁) / (L - (f₁ + f₂))
This is the image of the rays that enter parallel to the first surface