A businessman sells 60% of his wares at a profit of 15%. The remainder is sold at a loss of 25%. What is the percentage profit or loss on the sale of the whole wares?

Let the total cost price of the wares be $C$.
The cost price of 60% of the wares is $0.60C$.
The profit on this portion is 15%, so the selling price ($SP_1$) is:

$$SP_1 = 0.60C \times (1 + 0.15) = 0.60C \times 1.15 = 0.69C$$

The remainder is $100\% - 60\% = 40\%$ of the wares.
The cost price of this portion is $0.40C$.
The loss on this portion is 25%, so the selling price ($SP_2$) is:

$$SP_2 = 0.40C \times (1 - 0.25) = 0.40C \times 0.75 = 0.30C$$

The total selling price ($SP_{total}$) is the sum of $SP_1$ and $SP_2$:

$$SP_{total} = 0.69C + 0.30C = 0.99C$$

The total cost price is $C$.
Since $SP_{total} = 0.99C$ is less than $C$, there is an overall loss.

The total loss is $C - 0.99C = 0.01C$.
The percentage loss is:

$$\text{Percentage Loss} = \frac{\text{Loss}}{\text{Total Cost Price}} \times 100\%$$ $$\text{Percentage Loss} = \frac{0.01C}{C} \times 100\% = 0.01 \times 100\% = 1\%$$

The percentage profit or loss on the sale of the whole wares is a 1% loss.

To find $\vec{AB}$, subtract the coordinates of A from B:

$$\vec{AB} = B - A = (2 - (-2), -1 - 3) = (2 + 2, -1 - 3) = (4, -4)$$

To find $\vec{DC}$, subtract the coordinates of D from C:

$$\vec{DC} = C - D = (5 - m, 0 - n) = (5 - m, -n)$$

So, $\vec{AB} = \boxed{(4, -4)}$ and $\vec{DC} = \boxed{(5 - m, -n)}$.

For a parallelogram ABCD, $\vec{AB} = \vec{DC}$.
Equating the components of the vectors from Step 1 and Step 2:

$$(4, -4) = (5 - m, -n)$$

Equating the x-components:

$$4 = 5 - m \implies m = 5 - 4 \implies m = 1$$

Equating the y-components:

$$-4 = -n \implies n = 4$$

The coordinates of D are $\boxed{(1, 4)}$.

The coordinates are $D(1, 4)$ and $B(2, -1)$.
The distance formula is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

$$|DB| = \sqrt{(2 - 1)^2 + (-1 - 4)^2}$$ $$|DB| = \sqrt{(1)^2 + (-5)^2}$$ $$|DB| = \sqrt{1 + 25}$$ $$|DB| = \sqrt{26}$$

Calculating the numerical value and rounding to one decimal place:

$$|DB| \approx 5.0990195...$$ $$|DB| \approx \boxed{5.1}$$