Solve the inequalities and represent your answer on a number line.

In the right-angled triangle ABC, with the right angle at B:

$$AC^2 = AB^2 + BC^2$$ $$AC^2 = (12 \text{ cm})^2 + (5 \text{ cm})^2$$ $$AC^2 = 144 \text{ cm}^2 + 25 \text{ cm}^2$$ $$AC^2 = 169 \text{ cm}^2$$ $$AC = \sqrt{169 \text{ cm}^2}$$ $$AC = 13 \text{ cm}$$

The length of AC is $\boxed{\text{13 cm}}$.

From the triangle, $\theta$ is the angle at C.
First, find $\tan \theta$:

$$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{12}{5}$$

Now, substitute this into the expression:

$$1 + \tan^2 \theta = 1 + \left(\frac{12}{5}\right)^2$$ $$1 + \tan^2 \theta = 1 + \frac{144}{25}$$ $$1 + \tan^2 \theta = \frac{25}{25} + \frac{144}{25}$$ $$1 + \tan^2 \theta = \frac{169}{25}$$

The value of $(1 + \tan^2 \theta)$ is $\boxed{\frac{169}{25}}$.

The area of a right-angled triangle is given by:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Area} = \frac{1}{2} \times BC \times AB$$ $$\text{Area} = \frac{1}{2} \times 5 \text{ cm} \times 12 \text{ cm}$$ $$\text{Area} = \frac{1}{2} \times 60 \text{ cm}^2$$ $$\text{Area} = 30 \text{ cm}^2$$

The area of $\triangle ABC$ is $\boxed{\text{30 cm}^2}$.

From the triangle, $\theta$ is the angle at C.

$$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}$$

Using the value of AC from Step 1:

$$\sin \theta = \frac{12 \text{ cm}}{13 \text{ cm}}$$ $$\sin \theta = \frac{12}{13}$$

The value of $\sin \theta$ is $\boxed{\frac{12}{13}}$.