Find the value of x using the Pythagorean theorem.

In the right-angled triangle ABC, with the right angle at C, we have:

$$AC^2 + BC^2 = AB^2$$

Substitute the given values: $AC = x$, $BC = 8$ cm, $AB = 12$ cm.

$$x^2 + 8^2 = 12^2$$ $$x^2 + 64 = 144$$

Subtract 64 from both sides:

$$x^2 = 144 - 64$$ $$x^2 = 80$$

Take the square root of both sides:

$$x = \sqrt{80}$$

Simplify the square root:

$$x = \sqrt{16 \times 5}$$ $$x = 4\sqrt{5}$$

Calculate the numerical value of $x$:

$$x \approx 4 \times 2.2360679$$ $$x \approx 8.944 \text{ cm}$$

Rounding to one decimal place, $x \approx 8.9$ cm.

We can use the sine function, which relates the opposite side to the hypotenuse for angle $\theta$:

$$\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AB}$$

Substitute the given values:

$$\sin(\theta) = \frac{8}{12}$$ $$\sin(\theta) = \frac{2}{3}$$

To find $\theta$, take the inverse sine (arcsin) of $\frac{2}{3}$:

$$\theta = \arcsin\left(\frac{2}{3}\right)$$ $$\theta \approx \arcsin(0.6667)$$ $$\theta \approx 41.81^\circ$$

Rounding to one decimal place, $\theta \approx 41.8^\circ$.

Comparing our calculated values with the given options:
$x \approx 8.9$ cm and $\theta \approx 41.8^\circ$.
This matches option C.

The final answer is $\text{C. } 8.9\text{cm, } 41.8^\circ$.