Solutions

a)

$$C_{A\cap B}=C_A C_B.$$$$C_{A\cup B}=C_A+C_B-C_A C_B$$$$C_{R-A}=C_{R\cap \overline{A}}=C_{R}C_{\overline{A}}=C_{R}(1-C_A).$$

b) Define $A=\{x:f(x)=1\}$. Then, for $x\in A$, we have

$$f(x)=1=C_A(x).$$

Inversely, for $x\notin A$, we have

$$x\in \overline{A}=\{x:f(x)=0\},$$

so

$$f(x)=0=C_A(x).$$

Hence, $A$ is a set such that $f=C_A$.

c) In the forward direction $(\Rightarrow)$, suppose $f^2=f$. Then, simple algebraic manipulations yield

$$\begin{align*} f(x)^2-f(x)=0 &\implies f(x)(f(x)-1)=0 \\ &\implies f(x)=0 \lor f(x)=1. \end{align*}$$

Hence, explained in words, $f(x)=0$ or $1$ for each $x$. By what we proved in b), there is therefore a set $A$ such that $f=C_A$.

In the backward direction $(\Leftarrow)$, suppose $f=C_A$. Then, the range of $f$ is $\{0,1\}$. We also have that $0^2=0$ and $1^2=1$. Thus, $f=f^2$.