Ta có:\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)\left(x-3\right)\left(x+7\right)=297\)

\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+4x-21\right)=297\)

Đặt \(x^2+4x-5=t\) thì \(t\left(t-16\right)=297\)

\(\Leftrightarrow t^2-16t-297=0\Leftrightarrow t^2-27t+11t-297=0\)

\(\Leftrightarrow t\left(t-27\right)+11\left(t-27\right)=0\Leftrightarrow\left(t+11\right)\left(t-27\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=-11\\t=27\end{cases}}\)

Với \(t=-11\) thì \(x^2+4x-5=-11\Leftrightarrow x^2+4x+6=0\Leftrightarrow x^2+4x+4+2=0\)

   \(\Leftrightarrow\left(x+2\right)^2+2=0\)(vô lí)

Với \(t=27\) thì \(x^2+4x-5=27\Leftrightarrow x^2+4x-32=0\Leftrightarrow x^2-4x+8x-32=0\)

  \(\Leftrightarrow x\left(x-4\right)+8\left(x-4\right)=0\Leftrightarrow\left(x+8\right)\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-8\\x=4\end{cases}}\)

Tập nghiệm của pt \(S=\left\{-8,4\right\}\)
b) 

Ta có: \(\left(x^2+2x+3\right)^2-9\left(x^2+2x+3\right)+18=0\)

\(\Leftrightarrow\left(x^2+2x+3\right)^2-3\left(x^2+2x+3\right)-6\left(x^2+2x+3\right)+18=0\)

\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+3-3\right)-6\left(x^2+2x+3-3\right)=0\)

\(\Leftrightarrow\left(x^2+2x+3-6\right)\left(x^2+2x\right)=0\)

\(\Leftrightarrow\left(x^2+2x-3\right)\cdot x\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left(x^2+2x+1-4\right)\cdot x\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)\cdot x\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\cdot x\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=0\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{1;-3;0;-2\right\}\)