Given,
\(\begin{aligned}
& \frac{8}{(x+3)^2(x-2)}=\frac{A x+B}{(x+3)^2}+\frac{C}{x-2} \\
& \Rightarrow \frac{8}{(x+3)^2(x-2)}=\frac{(A x+B)(x-2)+C(x+3)^2}{(x+3)^2(x-4)} \\
& \Rightarrow \quad 8=(A x+B)(x-2)+C(x+3)^2 \\
& \text {At } x=2 \\
& \Rightarrow \quad 8=0+C(5)^2 \\
& \Rightarrow \quad 8=25 C \Rightarrow C=\frac{8}{25} \\
& \text {At } x=0 \text {, } \\
& \Rightarrow \quad 8=(B)(-2)+C(9) \\
& \Rightarrow \quad 8=-2 B+9 C \\
& \Rightarrow \quad 8=-2 B+\frac{72}{25} \quad\left(\because C=\frac{8}{25}\right) \\
& \Rightarrow \quad 2 B=\frac{72}{25}-8 \\
& \Rightarrow \quad 2 B=\frac{72-200}{25}=\frac{-128}{25} \\
& \Rightarrow \quad B=\frac{-64}{25} \\
& \text {At } \quad x=1 \text {, } \\
& \Rightarrow \quad 8=(A+B)(-1)+C(4)^2 \\
& \Rightarrow \quad 8=-A-B+16 C \\
& \Rightarrow \quad 8=-A+\frac{64}{25}+\frac{128}{25} \\
& \Rightarrow \quad A=\frac{192}{25}-8 \Rightarrow A=\frac{-8}{25}
\end{aligned}\)
Now, \(25(B+8 C-A)=25\left(\frac{-64}{25}+\frac{64}{25}+\frac{8}{25}\right)\)
\(=25 \times \frac{8}{25}=8\)