1. The problem asks to find the scale factor between triangles $\triangle ABC$ and $\triangle ADE$, then use it to find the perimeter of $\triangle ABC$ and the area of $\triangle ADE$.\n\n2. Given that $\triangle ABC \sim \triangle ADE$, the scale factor from $\triangle ADE$ to $\triangle ABC$ is the ratio of corresponding sides. From the problem, the scale factor is $\frac{7}{4}$.\n\n3. To find the perimeter of $\triangle ABC$, use the scale factor. The perimeter scales by the same factor as the sides. Given the perimeter of $\triangle ADE$ is 42, the perimeter of $\triangle ABC$ is:\n$$\text{Perimeter}_{ABC} = \text{Perimeter}_{ADE} \times \frac{7}{4} = 42 \times \frac{7}{4}$$\n\n4. Simplify the multiplication:\n$$42 \times \frac{7}{4} = \frac{42 \times 7}{4} = \frac{294}{4}$$\n\n5. Simplify the fraction by dividing numerator and denominator by 2:\n$$\frac{\cancel{294}^{147}}{\cancel{4}^2} = \frac{147}{2} = 73.5$$\n\n6. So, the perimeter of $\triangle ABC$ is 73.5 units.\n\n7. To find the area of $\triangle ADE$, recall that areas scale by the square of the scale factor. Since the scale factor from $\triangle ADE$ to $\triangle ABC$ is $\frac{7}{4}$, the scale factor from $\triangle ABC$ to $\triangle ADE$ is the reciprocal $\frac{4}{7}$.\n\n8. The area scale factor is then:\n$$\left(\frac{4}{7}\right)^2 = \frac{16}{49}$$\n\n9. Given the area of $\triangle ABC$ is 64, the area of $\triangle ADE$ is:\n$$64 \times \frac{16}{49} = \frac{1024}{49}$$\n\n10. Simplify the fraction if possible. Since 1024 and 49 share no common factors, the area is:\n$$\frac{1024}{49} \approx 20.90$$\n\n**Final answers:**\n- Scale factor: $\frac{7}{4}$\n- Perimeter of $\triangle ABC$: 73.5\n- Area of $\triangle ADE$: $\frac{1024}{49}$ or approximately 20.90