Answer:
a)    F₁ = 267.3 N,   N₁ = 1300 N,  b)    μ = 0.324
Explanation:
For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces
            
let's use subscript 1 for the ladder and 2 for the firefighter
             ∑ τ = 0
           -W₁ x₁ - W₂ x₂ + N₁ y = 0
            N₁ =  (1)
          (1)
the center of mass of the ladder is at its geometric center,
 d = L / 2 = 15/2 = 7.5 m
          cos 60 = x₁ / d₁
          x₁ = d₁ cos 60
          x₁ = 7.5 cos 60
          x₁ = 3.75 m
for the firefighter d₂ = 4 m
          cos 60 = x₂ / d₂
          x₂ = d₂ cos 60
           x₂ = 4 cos 60 = 2 m
for the fulcrum d₃ = 15 m
          sin 60 = y / d₃
          y = d₃ sin 60
          y = 15 sin 60
          y = 13 m
 we look for the Normal by substituting in equation 1
          N₂ =  
          N₂ = 267.3 N
now let's use the translational equilibrium relations
  X axis
            F₁ - N₂ = 0
            F₁ = N₂
            F₁ = 267.3 N
Axis y
           N₁ - W₁ -W₂ = 0
           N₁ = W₁ + W₂
           N₁ = 500 + 800
           N₁ = 1300 N
b) for this case change the firefighter's distance d₂ = 9 m
           x₂ = 9 cos 60
           x₂ = 4.5 m
we substitute in 1
           N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}  
           N₂ = 421.15 N
of the translational equilibrium equation on the x-axis
           fr = F₁ = N₂
           fr = 421.15 N
friction force has the expression
           fr = μ N
in this case the reaction of the Earth to the support of the ladder is N1 = 1300N
           μ = fr / N₁
           μ = 421.15 / 1300
           μ = 0.324