5. 6 1) (5-b)/(3-a) - (16-b^2)/(a^2-6a+9) * (3-a)/(b-4),

1) $$(\frac{5-b}{3-a} - \frac{16-b^2}{a^2-6a+9} \cdot \frac{3-a}{b-4}) = \frac{5-b}{3-a} - \frac{(4-b)(4+b)}{(a-3)^2} \cdot \frac{3-a}{b-4} = \frac{5-b}{3-a} - \frac{-(b-4)(4+b)}{(a-3)^2} \cdot \frac{3-a}{b-4} = \frac{5-b}{3-a} - \frac{-(4+b)}{a-3} \cdot \frac{3-a}{1} = \frac{5-b}{3-a} - (-(4+b)) \cdot \frac{-(a-3)}{a-3} = \frac{5-b}{3-a} - (4+b) = \frac{5-b - (4+b)(3-a)}{3-a} = \frac{5-b - (12 - 4a + 3b - ab)}{3-a} = \frac{5 - b - 12 + 4a - 3b + ab}{3-a} = \frac{-7 - 4b + 4a + ab}{3-a}$$.

Ответ: $$\frac{-7 - 4b + 4a + ab}{3-a}$$