Evaluate the numerator.

$$8-2+12 \times 9-4 \times 6$$

Perform multiplications first:

$$8-2+108-24$$

Perform additions and subtractions from left to right:

$$6+108-24$$ $$114-24$$ $$90$$

The numerator is 90.

$$56 \div 7(2)$$

Perform division first:

$$8(2)$$

Perform multiplication:

$$16$$

The denominator is 16.

$$\frac{90}{16}$$

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 2:

$$\frac{90 \div 2}{16 \div 2} = \frac{45}{8}$$

The fraction can also be expressed as a mixed number:

$$5\frac{5}{8}$$

The question asks to evaluate, so the simplified fraction is acceptable.

The final answer is $\boxed{\frac{45}{8}}$.

The dimensions are 840 m and 396 m. The side length of the largest possible square plot will be the GCD of these two numbers.
Using prime factorization:

$$840 = 2^3 \times 3 \times 5 \times 7$$ $$396 = 2^2 \times 3^2 \times 11$$

The GCD is the product of the lowest powers of common prime factors:

$$\text{GCD}(840, 396) = 2^2 \times 3 = 4 \times 3 = 12$$

So, the side length of each square plot is 12 m.

Number of plots along the 840 m side:

$$\frac{840 \text{ m}}{12 \text{ m/plot}} = 70 \text{ plots}$$

Number of plots along the 396 m side:

$$\frac{396 \text{ m}}{12 \text{ m/plot}} = 33 \text{ plots}$$

Total number of plots = (plots along length) $\times$ (plots along width)

$$70 \times 33 = 2310$$

The final answer is $\boxed{2310}$.

The distance OS is the radius ($R$) of the circumcircle. According to the sine rule, for any triangle with sides $a, b, c$ and opposite angles $A, B, C$, the following relationship holds:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$

In triangle RSQ, we are given side $QR = 12$ cm and its opposite angle $\angle RSQ = 50^\circ$.

$$\frac{QR}{\sin(\angle RSQ)} = 2R$$ $$\frac{12 \text{ cm}}{\sin(50^\circ)} = 2R$$

Solve for $R$:

$$R = \frac{12 \text{ cm}}{2 \sin(50^\circ)}$$ $$R = \frac{6 \text{ cm}}{\sin(50^\circ)}$$

Using a calculator for $\sin(50^\circ)$:

$$\sin(50^\circ) \approx 0.76604444$$

Now, calculate $R$:

$$R = \frac{6}{0.76604444} \approx 7.832409 \text{ cm}$$

Rounding to 2 decimal places:

$$R \approx 7.83 \text{ cm}$$

The distance OS is $R$.

The final answer is $\boxed{7.83 \text{ cm}}$.