Solution:

We know $X_3(k)=X_1(k) X_2(k)$

$ X_1(k)=\sum_{n=0}^{N-1} x_1(n) e^{-32 \pi k n / N} \quad k=0,1, \ldots N-1 $

Given $x_1(n)=\{1,1,2,1\}$ and $N=4$

$ \begin{aligned} & X_1(0)=\sum_{n=0}^3 x_1(n)=1+1+2+1=5 \\ & X_1(1)=\sum_{n=0}^3 x_1(n) e^{-j \pi n / 2}=1-j-2+j=-1 \\ & X_1(2)=\sum_{n=0}^3 x_1(n) e^{-j \pi n}=1-1+2-1=1 \\ & X_1(3)=\sum_{n=0}^3 x_1(n) e^{-j 3 \pi n / 2}=1+j-2-j=-1 \\ & X_1(k)=(5,-1,1,-1) \\ & X_2(k)=\sum_{n=0}^{N-1} x_2(n) e^{-j 2 \pi n k / N} \quad k=0,1, \ldots N-1 \end{aligned} $

$ \begin{aligned} & X_2(0)=\sum_{n=0}^3 x_2(n)=1+2+3+4=10 \\ & X_2(1)=\sum_{n=0}^3 x_2(n) e^{-j \pi n / 2}=1+2(-j)+3(-1)+4(j)=-2+j 2 \\ & X_2(2)=\sum_{n=0}^3 x_2(n) e^{-j \pi n}=1+2(-1)+3(1)+4(-1)=-2 \\ & X_2(3)=\sum_{n=0}^3 x_1(n) e^{-j 3 \pi n / 2}=1+2(j)+3(-1)+4(-j)=-2-j 2 \end{aligned} $

$ \begin{aligned} X_2(k) & =\{10,-2+j 2,-2,-2,-j 2\} \\ X_3(k) & =X_1(k) X_2(k)=\{50,2-j 2,-2,2+j 2\} \\ x_3(n) & =\frac{1}{N} \sum_{k=0}^{N-1} X_3(k) e^{j 2 \pi n k / N} \quad n=0,1, \ldots N-1 \\ x_3(0) & =\frac{1}{4} \sum_{k=0}^3 X_3(k)=\frac{1}{4}(50+2-j 2-2+2+j 2)=13 \\ x_3(1) & =\frac{1}{4}\left[\sum_{k=0}^4 X_3(k) e^{j \pi k / 2}\right] \\ & =\frac{1}{4}[50+(2-j 2) j+(-2)(-1)+(2+j 2)(-j)]=14 \\ x_3(2) & =\frac{1}{4}\left[\sum_{k=0}^4 X_3(k) e^{j \pi k}\right] \\ & =\frac{1}{4}[50+(2-j 2)(-j)+(-2)(-1)+(2+j 2)(j)]=12 \end{aligned} $

$ x_3(n)=\{13,14,11,12\} $