4) Simplify the expression: $(\frac{8b}{b+7} - \frac{15b}{b^2+14b+49}) : (\frac{8b+41}{b^2-49} + \frac{7b-49}{b+7})$

Let's simplify the first part of the expression:

$$\frac{8b}{b+7} - \frac{15b}{b^2+14b+49} = \frac{8b}{b+7} - \frac{15b}{(b+7)^2}$$

Find a common denominator which is $(b+7)^2$:

$$\frac{8b(b+7)}{(b+7)^2} - \frac{15b}{(b+7)^2} = \frac{8b^2 + 56b - 15b}{(b+7)^2} = \frac{8b^2 + 41b}{(b+7)^2}$$

Now let's simplify the second part of the expression:

$$\frac{8b+41}{b^2-49} + \frac{7b-49}{b+7} = \frac{8b+41}{(b-7)(b+7)} + \frac{7(b-7)}{b+7}$$

Find a common denominator which is $(b-7)(b+7)$:

$$\frac{8b+41}{(b-7)(b+7)} + \frac{7(b-7)(b+7)}{(b+7)(b-7)} = \frac{8b+41 + 7(b^2-49)}{(b-7)(b+7)} = \frac{8b+41+7b^2-343}{(b-7)(b+7)} = \frac{7b^2+8b-302}{(b-7)(b+7)}$$

So, the original expression becomes:

$$\frac{8b^2 + 41b}{(b+7)^2} : \frac{7b^2+8b-302}{(b-7)(b+7)} = \frac{b(8b+41)}{(b+7)^2} \cdot \frac{(b-7)(b+7)}{7b^2+8b-302} = \frac{b(8b+41)(b-7)}{(b+7)(7b^2+8b-302)}$$

Thus, we have:

$$\frac{b(8b+41)(b-7)}{(b+7)(7b^2+8b-302)}$$

Final Answer: $\frac{b(8b+41)(b-7)}{(b+7)(7b^2+8b-302)}$