$\tau_{max} = \dfrac{16T}{\pi D^3}$
For the 2-ft segment:
$\tau_{max2} = \dfrac{16(600)(12)}{\pi (2^3)} = 4583.66 \, \text{psi}$ answer
For the 3-ft segment:
$\tau_{max3} = \dfrac{16(800)(12)}{\pi (2^3)} = 6111.55 \, \text{psi}$ answer
Angle of twist
$\theta = \dfrac{TL}{JG}$
$\theta = \dfrac{1}{JG} \Sigma TL$
$\theta = \dfrac{1}{\frac{1}{31}\pi (2^4)(4 \times 10^6)} [ \, 600(2) + 800(3) \, ] \, (12^2)$
$\theta = 0.0825 \, \text{rad}$
$\theta = 4.73^\circ$ answer