$\theta_{mid} = 0$
$\theta_{uniform\,\,load} -\theta_{moment\,\,at\,\,midspan} = 0$
$\theta_1 - \theta_2 = 0$
See Case 3 and Case 5 of superposition method
$\dfrac{100(4^3)}{6EI} - \dfrac{M_{mid}(8)}{EI} = 0$
$8M_{mid} = 1066.67$
$M_{mid} = 133.33 \, \text{ ft}\cdot\text{lb}$
Moment at the fixed support (end moment):
$M = M_{mid} - 100(4)(2)$
$M = 133.33 - 800$
$M = -666.67 \, \text{ ft}\cdot\text{lb}$ answer
By symmetry, maximum deflection will occur at the midspan
$\delta_{max} = \delta_1 + 4\theta_1 - \delta_2$
$\delta_{max} = \dfrac{100(4^4)}{8EI} + 4 \left[ \dfrac{100(4^3)}{6EI} \right] - \dfrac{M_{mid}(8^2)}{2EI}$
$EI\delta_{max} = \dfrac{100(256)}{8} + \dfrac{4(100)(64)}{6} - \dfrac{133.33(64)}{2}$
$EI\delta_{max} = 3200 + 4266.67 - 4266.67$
$EI\delta_{max} = 3200 \, \text{ lb}\cdot\text{ft}^3$ answer
Note:
The fixed reactions are equal to $R = 100(4) = 400 \, \text{ lb}$. In the FBD of half left of the beam, $R$ equates to the uniform load; this made the vertical shear at the midspan equal to zero.
