Olá, Dan.
[tex](a)\ (i)\ D(f\circ f)=\{x\in\mathbb{R}|\frac{1}{x+1}\neq -1\}=\{x\in\mathbb{R}|1\neq -x-1\}=\\\\=\boxed{\{x\in\mathbb{R}|x\neq -2,x\neq-1\}=\mathbb{R}-\{-2,-1\}}[/tex]
[tex]f\circ f(x)=\frac1{1+\frac{1}{1+x}}=\frac{1}{\frac{1+x+1}{1+x}}\Rightarrow\boxed{f(x)=\frac{1+x}{2+x}}[/tex]
[tex](b)\ \frac1{1+x}\neq0\ (\forall x\in\mathbb{R}\text{ satisfaz, desde que }x\neq-1)\\\\ \frac{1}{1+x} \neq 1 \Rightarrow 1\neq1+x\Rightarrow x\neq0,x\neq-1\\\\ \frac{1}{1+x} \neq 2 \Rightarrow 1\neq2+2x\Rightarrow 2x\neq-1 \Rightarrow x\neq-\frac12,x\neq-1\\\\ \therefore \boxed{D(g\circ f)=\left\{x\in\mathbb{R}|x\neq0,x\neq-\frac12,x\neq-1\right\}=\mathbb{R}-\left\{0,-\frac12,-1\right\}}[/tex]
[tex](c)[/tex] O domínio [tex]A\bigcap B[/tex] satisfaz tanto a função [tex]f[/tex] quanto a função [tex]g,[/tex] pois:
[tex]x\in A\bigcap B \Rightarrow \begin{cases}x\in A \Rightarrow x\in D(f)\\x\in B \Rightarrow x\in D(g) \end{cases} \Rightarrow x\in D(f) \bigcap D(g) \Rightarrow[/tex]
f + g e f.g estão bem definidas
[tex]D(f+g)=D(f\cdot g)=(\mathbb{R}-\{-1\}) \bigcap\ (\mathbb{R}-\{0,1,2\}) \Rightarrow \\\\ \boxed{D(f+g)=D(f\cdot g)=\mathbb{R}-\{-1,0,1,2\}}[/tex]
