$8.28 \log _{\pi}(x+27)-\log _{\pi}(16-2 x)<\log _{\pi} x$.
8.28 The given inequality is equivalent to the system

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x + 27 > 0 , } \\
{ 16 - 2 x > 0 , } \\
{ x > 0 , } \\
{ \operatorname { log } _ { \pi } \frac { x + 27 } { 16 - 2 x } - 27 \\
x > 0, \\
\frac{x + 27}{16 - 2 x} > 0
\end{array}\right.\right.
\end{aligned}
$$

Solving it, we find $3 < x < 4.5$.

Answer: $(3 ; 4.5)$.