a)\(\left(x-1\right)\left(3-x\right)>0\)

\(\Leftrightarrow-\left(x-1\right)\left(x-3\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)< 0\)

\(\Rightarrow\)\(x-1;x-3\) are the numbers opposite the sign

We have: \(x-1>x-3\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}x-1>0\\x-3< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\)

b)\(xy=x+y\)

\(\Rightarrow xy-x-y+1=1\)

\(\Rightarrow x\left(y-1\right)-\left(y-1\right)=1\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=1\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(;\left\{{}\begin{matrix}x-1=-1\\y-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y=2\\x=y=0\end{matrix}\right.\)

a) \(\left(x-1\right)\left(3-x\right)>0\),so x - 1 and 3 - x are 2 sign numbers

1^st case : \(\left\{{}\begin{matrix}x-1>0\\3-x>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\)\(\Rightarrow1< x< 3\)

2^nd case : \(\left\{{}\begin{matrix}x-1< 0\\3-x< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 1\\x>3\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Hence, 1 < x < 3

b) \(xy=x+y\Leftrightarrow xy-x-y=0\)\(\Leftrightarrow x\left(y-1\right)-\left(y-1\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)=1\)\(\Rightarrow x-1=\dfrac{1}{y-1}\Rightarrow x=\dfrac{y}{y-1}\)\(\left(y\ne1\right)\)

If you give \(x\in Z\),the answer of question a) is x = 2 and I'll solve the question b) again