Answer:
a)  θ₁ = 0.487º
, b)   t = 0.400 s
,        x = 11.73 ft
Explanation:
For this exercise let's use the projectile launch relationships.
The initial height is I = 5 ft and the final height y = 4 ft
             y = y₀ +  t - ½ g t²
 t - ½ g t²
The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft
             x = v₀ₓ t
             t = x / v₀ₓ
We replace
              y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²
              v_{oy} = v₀ sin θ
              v₀ₓ = vo cos θ
              
              y –y₀ = x tan θ - ½ g x² / v₀² cos² θ
                 5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)
                 1 = 120 tan θ - 0.0213 sec² θ
Let's use the trigonometry relationship
                Sec² θ = 1 - tan² θ
                  1 = 120 tan θ - 0.0213 (1 –tan²θ)
                  0.0213 tan²θ + 120 tanθ -1.0213 = 0
                  
We change variables
           u = tan θ
           u² + 5633.8 u - 48.03 = 0
We solve the second degree equation
           u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2
           u = [- 5633.8 ± 5633.82] / 2
            u₁ = 0.0085
            u₂= -5633.81
            u = tan θ
            θ = tan⁻¹ u
For u₁
            θ₁ = tan⁻¹ 0.0085
            θ₁ = 0.487º
For u₂
            θ₂ = -89.99º
The launch angle must be 0.487º
b) let's look for the time it takes for the arrow to arrive
          x = v₀ₓ t
          t = x / v₀ cos θ
          
          t = 120 / (300 cos 0.487)
          t = 0.400 s
The deer must be at a distance of
            v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s
            x = v t
            x = 29.33 0.4
            x = 11.73 ft