1. Stating the problem: We have triangle OMN with vectors $\overrightarrow{OM} = a$ and $\overrightarrow{ON} = b$. Point R lies on MN such that $MR : RN = 3 : 2$. We need to prove $\overrightarrow{OR} = \frac{2}{5}a + \frac{3}{5}b$. 2. To find $\overrightarrow{OR}$, express $\overrightarrow{OR}$ in terms of $\overrightarrow{OM}$ and $\overrightarrow{ON}$. Note that $\overrightarrow{OR} = \overrightarrow{OM} + \overrightarrow{MR}$. 3. Since $R$ divides $MN$ in ratio $3:2$, vector $\overrightarrow{MR} = \frac{3}{5}\overrightarrow{MN}$. 4. Vector $\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = b - a$. 5. Therefore, $\overrightarrow{OR} = a + \frac{3}{5}(b - a) = a + \frac{3}{5}b - \frac{3}{5}a = \frac{2}{5}a + \frac{3}{5}b$. 6. Hence, part (i) is shown: $\boxed{\overrightarrow{OR} = \frac{2}{5}a + \frac{3}{5}b}$. 7. For part (ii)(a), given $\overrightarrow{NT} = 4a + kb$ and $\overrightarrow{OT} = c \overrightarrow{OR} = c \left( \frac{2}{5}a + \frac{3}{5}b \right)$. 8. Express $\overrightarrow{OT}$ in terms of $\overrightarrow{ON}$ and $\overrightarrow{NT}$: $$\overrightarrow{OT} = \overrightarrow{ON} + \overrightarrow{NT} = b + 4a + kb = 4a + (k + 1)b.$$ 9. Equate to $c \overrightarrow{OR}$: $$4a + (k + 1)b = c \left( \frac{2}{5}a + \frac{3}{5}b \right) = \frac{2c}{5}a + \frac{3c}{5}b.$$ 10. Equate coefficients of $a$ and $b$: $$4 = \frac{2c}{5} \implies c = 10,$$ $$k + 1 = \frac{3c}{5} = \frac{3 \times 10}{5} = 6 \implies k = 5.$$ 11. So, $\boxed{k = 5}$ and $\boxed{c = 10}$. 12. For part (ii)(b), find $\overrightarrow{MT}$: $$\overrightarrow{MT} = \overrightarrow{OT} - \overrightarrow{OM} = (4a + (k+1)b) - a = 4a + 6b - a = 3a + 6b.$$ 13. Therefore, $$\boxed{\overrightarrow{MT} = 3a + 6b}.$$