1. \( \frac{x^{2}+10 x+16}{5 x-10} \cdot \frac{x-2}{x^{2}+9 x+8} \)

To simplify the expression \( \frac{x^{2}+10 x+16}{5 x-10} \cdot \frac{x-2}{x^{2}+9 x+8} \), let's first factor the polynomials in the numerators and denominators. The first numerator \( x^{2}+10x+16 \) factors to \( (x+2)(x+8) \) since \( 2 \times 8 = 16 \) and \( 2 + 8 = 10 \). The denominator \( 5x-10 \) can be factored as \( 5(x-2) \). Next, look at the second numerator \( x-2 \), which is already factored, and the second denominator \( x^{2}+9x+8 \) factors to \( (x+1)(x+8) \) because \( 1 \times 8 = 8 \) and \( 1 + 8 = 9 \). Putting it all together, we have: \[ \frac{(x+2)(x+8)}{5(x-2)} \cdot \frac{x-2}{(x+1)(x+8)} \] Now, we can cancel the \( (x-2) \) terms in the numerator and denominator, as well as \( (x+8) \): \[ \frac{x+2}{5(x+1)} \] So, the simplified expression is: \[ \frac{x+2}{5(x+1)} \]