Solve for the marked sides in the given triangles. Assume all triangles are right-angled.

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:

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

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

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

$$x = \frac{80}{\frac{\sqrt{2}}{2}} = \frac{160}{\sqrt{2}} = \frac{160\sqrt{2}}{2} = 80\sqrt{2}$$ $$x \approx 80 \times 1.4142 = 113.136$$

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

In the left right-angled triangle, the hypotenuse is $8 \text{ cm}$ and the angle is $60^\circ$.

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

In the right right-angled triangle, the altitude is $h = 4\sqrt{3} \text{ cm}$ and the angle is $45^\circ$.

$$\sin(45^\circ) = \frac{h}{Hy}$$ $$Hy = \frac{h}{\sin(45^\circ)} = \frac{4\sqrt{3} \text{ cm}}{\frac{\sqrt{2}}{2}} = \frac{8\sqrt{3} \text{ cm}}{\sqrt{2}} = \frac{8\sqrt{3}\sqrt{2} \text{ cm}}{2} = 4\sqrt{6} \text{ cm}$$

First, find the base segment of the left triangle ($x_1$) and the right triangle ($x_2$).
For the left triangle:

$$\cos(60^\circ) = \frac{x_1}{8 \text{ cm}}$$ $$x_1 = 8 \text{ cm} \times \cos(60^\circ) = 8 \text{ cm} \times \frac{1}{2} = 4 \text{ cm}$$

For the right triangle:

$$\tan(45^\circ) = \frac{h}{x_2}$$ $$x_2 = \frac{h}{\tan(45^\circ)} = \frac{4\sqrt{3} \text{ cm}}{1} = 4\sqrt{3} \text{ cm}$$

The total base $Y$ is the sum of $x_1$ and $x_2$:

$$Y = x_1 + x_2 = 4 \text{ cm} + 4\sqrt{3} \text{ cm} = 4(1+\sqrt{3}) \text{ cm}$$

The marked sides are $Hy = \boxed{\mathbf{4\sqrt{6} \text{ cm}}}$ and $Y = \boxed{\mathbf{4(1+\sqrt{3}) \text{ cm}}}$.