Given that ABC DEF, the ratio of their corresponding sides is equal.

The corresponding sides are $AB$ and $DE$, and $AC$ and $DF$.

$$\frac{AB}{DE} = \frac{AC}{DF}$$

We have $AB = 8 \text{ cm}$, $AC = 6 \text{ cm}$, and $DF = 12 \text{ cm}$.

$$\frac{8 \text{ cm}}{DE} = \frac{6 \text{ cm}}{12 \text{ cm}}$$

Simplify the right side:

$$\frac{8}{DE} = \frac{1}{2}$$

Cross-multiply:

$$8 \times 2 = DE \times 1$$ $$16 = DE$$

Therefore, the length of side $DE$ is:

$$DE = \boxed{\text{16 cm}}$$

:
Given that the ratio of corresponding sides of two similar triangles is $3:5$. One side of the smaller triangle is $9 \text{ cm}$.

Let $s_S$ be the side of the smaller triangle and $s_L$ be the corresponding side of the larger triangle.

$$\frac{s_S}{s_L} = \frac{3}{5}$$

We are given $s_S = 9 \text{ cm}$.

$$\frac{9 \text{ cm}}{s_L} = \frac{3}{5}$$

Cross-multiply:

$$9 \times 5 = 3 \times s_L$$ $$45 = 3 s_L$$

Divide by 3:

$$s_L = \frac{45}{3}$$ $$s_L = 15 \text{ cm}$$

The corresponding side of the larger triangle is:

$$s_L = \boxed{\text{15 cm}}$$