Answer: Attached below is the missing detail and Mohr's circle.
i) б1 =  9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) Â -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
   = -20 / 0.785 * 5^2  = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
 = (16 * 250 ) / π * 5^3  = 010.19 Ksi
Calculate the Principal stress, maximum in-plane shear stress and average normal stress
Using Mohr's circle ( attached below )
i) principal stresses:
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
ii) maximum in-plane shear stress
ζ  = radius of Mohr's circle
  = 5.1 cm = 10.2 Ksi  ( Given that ; 1 cm = 2Ksi )
iii) average normal stress
 = 9.6 + ( - 10.7 ) / 2
 = -0.51Ksi