Let the mid points of the sides of a triangle ABC be ((5)/(2), 7) ,...

Let the vertices of the triangle $ABC$ be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

Let the given midpoints be of sides $BC$, $CA$, and $AB$ respectively:
Midpoint of $BC = \left(\dfrac{5}{2}, 7\right) \Rightarrow x_2 + x_3 = 5$ and $y_2 + y_3 = 14$
Midpoint of $CA = \left(\dfrac{5}{2}, 3\right) \Rightarrow x_3 + x_1 = 5$ and $y_3 + y_1 = 6$
Midpoint of $AB = (4, 5) \Rightarrow x_1 + x_2 = 8$ and $y_1 + y_2 = 10$

Adding the three equations for the $x$-coordinates:
$2(x_1 + x_2 + x_3) = 5 + 5 + 8 = 18$
$x_1 + x_2 + x_3 = 9$

Substituting the known sums:
$x_1 = 9 - 5 = 4$
$x_2 = 9 - 5 = 4$
$x_3 = 9 - 8 = 1$

Similarly, for $y$-coordinates:
$2(y_1 + y_2 + y_3) = 14 + 6 + 10 = 30$
$y_1 + y_2 + y_3 = 15$

$y_1 = 15 - 14 = 1$
$y_2 = 15 - 6 = 9$
$y_3 = 15 - 10 = 5$

So, the vertices are $A(4, 1)$, $B(4, 9)$, and $C(1, 5)$.

Lengths of the sides:
$a = BC = \sqrt{(4 - 1)^2 + (9 - 5)^2} = \sqrt{9 + 16} = 5$
$b = CA = \sqrt{(4 - 1)^2 + (1 - 5)^2} = \sqrt{9 + 16} = 5$
$c = AB = \sqrt{(4 - 4)^2 + (9 - 1)^2} = \sqrt{0 + 64} = 8$

Coordinates of the incentre $(h, k)$:
$I(x, y) = \left(\dfrac{ax_1 + bx_2 + cx_3}{a + b + c}, \dfrac{ay_1 + by_2 + cy_3}{a + b + c}\right)$

$h = \dfrac{5(4) + 5(4) + 8(1)}{5 + 5 + 8} = \dfrac{20 + 20 + 8}{18} = \dfrac{48}{18} = \dfrac{8}{3}$

$k = \dfrac{5(1) + 5(9) + 8(5)}{5 + 5 + 8} = \dfrac{5 + 45 + 40}{18} = \dfrac{90}{18} = 5$

So, the incentre is $(h, k) = \left(\dfrac{8}{3}, 5\right)$.

Therefore:
$3h + k = 3 \cdot \dfrac{8}{3} + 5 = 8 + 5 = 13$

Hence, the correct option is $(3)\ 13$.