$$\begin{aligned} & \langle\mathrm{p}\rangle=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})}{4}=6 \\ & \overrightarrow{\mathrm{a}}=\left(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{~m}}=\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \\ & \overrightarrow{\mathrm{v}} \text { at } \mathrm{t}=4 \mathrm{sec}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \times 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \\ & P_{\text {ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13 \\ & \frac{\langle\mathrm{P}\rangle}{P_{\text {ins }}}=\frac{6}{13} \end{aligned}$$

Note: Given data is not matching.

$$\begin{aligned} & S=u t+\frac{1}{2} a t^2 \\ & S=0+\frac{1}{2} \frac{(2 \hat{i}+3 \hat{j})}{4}(4)^2=4 \hat{i}+6 \hat{\mathrm{j}} \end{aligned}$$

If $\overrightarrow{\mathrm{r}}_{\mathrm{i}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}$ then $\overrightarrow{\mathrm{r}}_{\mathrm{f}}=7 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$

But Final position given in the question is $(6,10)$.