Find the lengths of the marked sides x and y.

Here are the solutions for the marked sides:

a)

Step 1: Identify the trigonometric relationship.
In the right-angled triangle, we are given the angle $45^\circ$, the side opposite to it (80), and we need to find the hypotenuse ($x$). The sine function relates these:

$$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$

Step 2: Substitute the values and solve for $x$.

$$\sin(45^\circ) = \frac{80}{x}$$ $$x = \frac{80}{\sin(45^\circ)}$$

Since $\sin(45^\circ) = \frac{\sqrt{2}}{2}$:

$$x = \frac{80}{\frac{\sqrt{2}}{2}} = \frac{160}{\sqrt{2}}$$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$$x = \frac{160\sqrt{2}}{2} = 80\sqrt{2}$$

The value of $x$ is $\boxed{\mathbf{80\sqrt{2}}}$.

b)
This problem involves two right-angled triangles sharing a common altitude. Let $h$ be the common altitude.

Step 1: Calculate the altitude $h$ using the left triangle.
In the left right-angled triangle, the hypotenuse is $8 \text{ cm}$ and the angle is $60^\circ$.

$$\sin(60^\circ) = \frac{h}{8 \text{ cm}}$$ $$h = 8 \text{ cm} \times \sin(60^\circ) = 8 \text{ cm} \times \frac{\sqrt{3}}{2} = 4\sqrt{3} \text{ cm}$$

Step 2: Calculate the hypotenuse $Hy$ of the right triangle.
In the right right-angled triangle, the altitude is $h = 4\sqrt{3} \text{ cm}$ and the angle is $45^\circ$.

$$\sin(45^\circ) = \frac{h}{Hy}$$ $$$$