(4) The vector a = (7) and b = (3) are such that a + 2b = c. Find the values of m and n

The displacement vector $\vec{PQ}$ is found by subtracting the position vector of the starting point P ($\vec{OP}$) from the position vector of the ending point Q ($\vec{OQ}$).

$$\vec{PQ} = \vec{OQ} - \vec{OP}$$

To find the coordinates of P, we need to solve for $\vec{OP}$. Rearrange the equation:

$$\vec{OP} = \vec{OQ} - \vec{PQ}$$

Subtract the corresponding components:

$$\vec{OP} = \begin{pmatrix} -1 - 2 \\ 4 - 3 \end{pmatrix}$$

The position vector $\vec{OP} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ means that the coordinates of point P are $\boxed{(-3, 1)}$.

4) The vectors $a = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$, $b = \begin{pmatrix} m \\ -3 \end{pmatrix}$ and $c = \begin{pmatrix} 8 \\ n \end{pmatrix}$ are such that $a + 2b = c$. Find the values of m and n.

Multiply each component of vector $b$ by 2:

$$2 \begin{pmatrix} m \\ -3 \end{pmatrix} = \begin{pmatrix} 2m \\ -6 \end{pmatrix}$$

Substitute the result back and add the corresponding components:

$$\begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 2m \\ -6 \end{pmatrix} = \begin{pmatrix} 8 \\ n \end{pmatrix}$$ $$\begin{pmatrix} 2 + 2m \\ 5 + (-6) \end{pmatrix} = \begin{pmatrix} 8 \\ n \end{pmatrix}$$ $$\begin{pmatrix} 2 + 2m \\ -1 \end{pmatrix} = \begin{pmatrix} 8 \\ n \end{pmatrix}$$

For the vectors to be equal, their x-components must be equal and their y-components must be equal.

$$2 + 2m = 8$$ $$-1 = n$$

From the y-components, we get:

$$n = -1$$

From the x-components:

$$2m = 8 - 2$$ $$2m = 6$$ $$m = \frac{6}{2}$$ $$m = 3$$

The values are $\boxed{m = 3, n = -1}$.