Five geometrical sets and three mathematical tables cost sh. 2,816. Three geometrical sets and five mathematical tables cost sh. 3,360. Find the cost of one geometrical set and one mathematical table

Let $g$ be the cost of one geometrical set.
Let $m$ be the cost of one mathematical table.

From the problem statement, we have two equations:

1. Five geometrical sets and three mathematical tables cost sh. 2,816:

$$5g + 3m = 2816$$

2. Three geometrical sets and five mathematical tables cost sh. 3,360:

$$3g + 5m = 3360$$

$$\begin{bmatrix} 5 & 3 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} g \\ m \end{bmatrix} = \begin{bmatrix} 2816 \\ 3360 \end{bmatrix}$$

Here, $A = \begin{bmatrix} 5 & 3 \\ 3 & 5 \end{bmatrix}$, $X = \begin{bmatrix} g \\ m \end{bmatrix}$, and $B = \begin{bmatrix} 2816 \\ 3360 \end{bmatrix}$.

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is $\frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$$A^{-1} = \frac{1}{16} \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix}$$

$$\begin{bmatrix} g \\ m \end{bmatrix} = \frac{1}{16} \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} \begin{bmatrix} 2816 \\ 3360 \end{bmatrix}$$

Perform the matrix multiplication:

$$\begin{bmatrix} g \\ m \end{bmatrix} = \frac{1}{16} \begin{bmatrix} (5 \times 2816) + (-3 \times 3360) \\ (-3 \times 2816) + (5 \times 3360) \end{bmatrix}$$ $$\begin{bmatrix} g \\ m \end{bmatrix} = \frac{1}{16} \begin{bmatrix} 14080 - 10080 \\ -8448 + 16800 \end{bmatrix}$$ $$\begin{bmatrix} g \\ m \end{bmatrix} = \frac{1}{16} \begin{bmatrix} 4000 \\ 8352 \end{bmatrix}$$

$$g = \frac{4000}{16} = 250$$ $$m = \frac{8352}{16} = 522$$

The cost of one geometrical set is sh. 250.
The cost of one mathematical table is sh. 522.

The final answer is $\boxed{\text{Cost of one geometrical set = sh. 250, Cost of one mathematical table = sh. 522}}$.