The question must give the conditions of x.I'll let \(x\in Z\)

a) \(A=\dfrac{x+5}{x+2}=1+\dfrac{3}{x+2}\)

\(A\in Z\Rightarrow\dfrac{3}{x+2}\in Z\Rightarrow3⋮x+2\)\(\Rightarrow x+2\in\left\{\pm1;\pm3\right\}\)

\(\Rightarrow x\in\left\{-5;-3;-1;1\right\}\)

b) A gets the largest value when \(\dfrac{3}{x+2}\) gets the largest value

\(\Rightarrow x+2=1\) or x = -1,then \(A=1+\dfrac{3}{1}=4\)

So,x = -1