$\Sigma M_{\text{to the left of }B} = 0$

$3.5R_A - 2.5(3.5) \left( \dfrac{3.5}{2} \right) = 0$

$R_A = 4.375 ~ \text{kN}$ ← Answer for Part (1)

By symmetry:

$R_C = R_A$

$R_C = 4.375 ~ \text{kN}$

$\Sigma; F_V = 0$

$R_B + R_A + R_C = 7(2.5)$

$R_B = 8.75 ~ \text{kN}$ ← Answer for Part (2)

$\Delta = \Delta_\text{uniform load} - \Delta_{\text{reaction at }B}$

$\Delta = \dfrac{5wL^4}{384EI} - \dfrac{PL^3}{48EI}$

$\Delta = \dfrac{5(2.5)(7^4)(1000^4)}{384(2.8 \times 10^{11})} - \dfrac{8.75(7^3)(1000^4)}{48(2.8 \times 10^{11})}$

$\Delta = 55.83 ~ \text{mm}$ ← Answer for Part (3)