1. **State the problem:** Find the lengths of the major and minor axes of the ellipse given by the equation $$\frac{(x-4)^2}{25} + \frac{(y-3)^2}{9} = 1.$$\n\n2. **Recall the ellipse standard form and axis lengths:** The standard form of an ellipse centered at $(h,k)$ is $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,$$ where $a$ and $b$ are the semi-major and semi-minor axes respectively. The lengths of the major and minor axes are $2a$ and $2b$.\n\n3. **Identify $a^2$ and $b^2$ from the equation:** Here, $a^2 = 25$ and $b^2 = 9$. Since $a^2 > b^2$, the major axis is horizontal.\n\n4. **Calculate the lengths of the axes:**\n$$a = \sqrt{25} = 5,$$\n$$b = \sqrt{9} = 3.$$\nTherefore, the major axis length is $$2a = 2 \times 5 = 10,$$\nand the minor axis length is $$2b = 2 \times 3 = 6.$$\n\n5. **Final answer:** The length of the major axis is 10 and the length of the minor axis is 6.