$\Sigma \theta_{mid} = 0$
See Case 2 and Case 5 of superposition method:
$\dfrac{M_{mid}(\frac{3}{2}a)}{EI} - \dfrac{Pa^2}{2EI} = 0$
$\dfrac{3aM_{mid}}{2EI} = \dfrac{Pa^2}{2EI}$
$M_{mid} = \frac{1}{3}Pa$
Thus,
$M_{wall} = M_{mid} - Pa$
$M_{wall} = \frac{1}{3}Pa - Pa$
$M_{wall} = -\frac{2}{3}Pa$ answer
Again, see Case 2 and Case 5 of superposition method:
$\delta_{mid} = \dfrac{Pa^2}{6EI}[ \, 3(\frac{3}{2}a) - a \, ] - \dfrac{M_{mid}(\frac{3}{2}a)^2}{2EI}$
$\delta_{mid} = \dfrac{Pa^2}{6EI}(\frac{7}{2}a) - \dfrac{\frac{1}{3}Pa(\frac{9}{4}a^2)}{2EI}$
$\delta_{mid} = \dfrac{7Pa^3}{12EI} - \dfrac{3Pa^3}{8EI}$
$\delta_{mid} = \dfrac{5Pa^3}{24EI}$
$EI\delta_{mid} = \frac{5}{24}Pa^3$ answer
