$\Sigma M_{R2} = 0$
$5R_1 = 800 + 1200(1)$
$R_1 = 400 ~ \text{N}$
$\Sigma M_{R1} = 0$
$5R_2 + 800 = 1200(4)$
$R_2 = 800 ~ \text{N}$
$M_1 = M_4 = 0$
$M_2 = 2R_1 - 800 = 2(400) - 800 = 0$
$M_3 = 1R_2 = 800 ~ \text{N}\cdot\text{m}$
For the deflection at point 2 (consider the span 1-2-4)
$L_1 = 2 ~ \text{m}$ ← span 1 to 2
$L_2 = 3 ~ \text{m}$ ← span 2 to 4
$M_1 L_1 + 2M_2(L_1 + L_2) + M_4L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_4}{L_2} \right)$
$0 + 0 + 0 - \dfrac{-800}{2}[ \, 3(2^2) - 2^2 \, ] + \dfrac{1200(1)}{3}(3^2 - 1^2) = 6EI \left( \dfrac{h}{2} + \dfrac{h}{3} \right)$
$3200 + 3200 = 6EI \left( \dfrac{5h}{6} \right)$
$6400 = 5EI \, h$
$h = \dfrac{1280}{EI}$
The positive sign indicates that points 1 and 4 are above point 2
$\delta_2 = \dfrac{1280}{EI} ~ \text{ downward}$ answer
For the deflection at point 3 (consider the span 1-3-4)
$L_1 = 4 ~ \text{m}$ ← span 1 to 3
$L_2 = 1 ~ \text{m}$ ← span 3 to 4
$M_1 L_1 + 2M_3(L_1 + L_2) + M_4L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_1}{L_1} + \dfrac{h_4}{L_2} \right)$
$0 + 2(800)(4 + 1) + 0 - \dfrac{-800}{4}[ \, 3(2^2) - 4^2 \, ] + 0 = 6EI \left( \dfrac{h}{4} + \dfrac{h}{1} \right)$
$8000 - 800 = 6EI \left( \dfrac{5h}{4} \right)$
$7200 = \frac{15}{2}EI \, h$
$h = \dfrac{960}{EI}$
The positive sign indicates that points 1 and 4 are above point 3
$\delta_3 = \dfrac{960}{EI} ~ \text{ downward}$ answer
