$\dfrac{R_2 L^3}{48EI} = \dfrac{5wL^4}{384EI}$

$R_2 = \frac{5}{8}wL = \frac{5}{8}(21.5)(12)$

$R = 161.25 ~ \text{kN}$ ← *Answer for Part (3)*

$\Sigma F_v = 0$

$2R_1 + R_2 = 21.5(12)$

$2R_1 + 161.25 = 258$

$R_1 = 48.375 ~ \text{kN}$

$M_2 = 48.375(6) - 21.5(6)(3)$

$M_2 = -96.75 ~ \text{kN}\cdot\text{m}$ ← *Answer for Part (1)*

$V_x = 48.375 - 21.5x = 0$

$x = 2.25 ~ \text{m}$

$M_x = 48.375x - 21.5x\left( \dfrac{x}{2} \right)$

$M_x = 48.375(2.25) - 10.75(2.25^2)$

$M_x = 54.42 ~ \text{kN}\cdot\text{m}$ ← *Answer for Part (2)*