Let $(h, k)$ be the pole of the line $9 x+y-28=$ with respect to the circle $x^2+y^2-\frac{3}{2} x+\frac{5}{2} y-\frac{7}{2}=0$. Then, the equation of polar is
$$
\begin{aligned}
& \text { polar is } \\
& \qquad h x+k y-\frac{3}{4}(x+h)+\frac{5}{4}(y+k)-\frac{7}{2}=0 \\
& \Rightarrow x\left(h-\frac{3}{4}\right)+y\left(k+\frac{5}{4}\right)-\frac{3}{4} h+\frac{5}{4} k-\frac{7}{2}=0 \\
& \Rightarrow \quad x(4 h-3)+y(4 k+5)-3 h+5 k-14=0
\end{aligned}
$$
This equation and $9 x+y-28=0$ represent the same line.
$$
\begin{array}{lrr}
\therefore & \frac{4 h-3}{9}=\frac{4 k+5}{1}=\frac{-3 h+5 k-14}{-28}=\lambda \text { (say) } \\
\Rightarrow h= & \frac{3+9 \lambda}{4}, k=\frac{\lambda-5}{4},-3 h+5 k-14=-28 \lambda \\
\Rightarrow & -3\left(\frac{3+9 \lambda}{4}\right)+5\left(\frac{\lambda-5}{4}\right)-14=-28 \lambda \\
\Rightarrow & -9-27 \lambda+5 \lambda-25-56=-112 \lambda \\
\Rightarrow & -22 \lambda-90=-112 \lambda \\
\Rightarrow & & 90 \lambda=90 \Rightarrow \lambda=1
\end{array}
$$
Hence, the pole of the given line is $(3,-1)$.