We have, $A=\{1,2,3,4,5\}$ and $S=\{(x, y): x \in A, y \in A\}$

(i) The set of ordered pairs satisfying $x+y=5$ is,

$$\{(1,4),(2,3),(3,2),(4,1)\}$$

(ii) The set of ordered pairs satisfying $x+y<5$ is $\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)$.

(iii) The set of ordered pairs satisfying $x+y>8$ is $\{(4,5),(5,4),(5,5)\}$.

We have,

$\begin{aligned} R & =\left\{(x, y): x, y \in W, x^2+y^2=25\right\} \\ & =\{(0,5),(3,4),(4,3),(5,0)\}\end{aligned}$

$\begin{aligned} \text{Domain of }\quad R & =\text { Set of first element of ordered pairs in } R \\ & =\{0,3,4,5\}\end{aligned}$

$\begin{aligned} \text{Range of }\quad R & =\text { Set of second element of ordered pairs in } R \\ & =\{5,4,3,0,\}\end{aligned}$

We have, $\quad R_1=\{x, y) \mid y=2 x+7$, where $x \in R$ and $\left.-5 \leq x \leq 5\right\}$

Domain of $R_1=\{-5 \leq x \leq 5, x \in R\}$

$=[-5,5]$

$\begin{array}{ll}\because & y=2 x+7 \\ \text { When } x=-5 \text {, then } & y=2(-5)+7=-3 \\ \text { When } x=5 \text {, then } & y=2(5)+7=17\end{array}$

$$\begin{aligned} & \therefore \quad \text { Range of } R_1=\{-3 \leq y \leq 17, y \in R\} \\ & =[-3,17] \end{aligned}$$

We have, $\mathrm{R}_2=\{(x, y)\} x$ and $y$ are integers and $\left.x^2+y^2=64\right\}$

Since, 64 is the sum of squares of 0 and $\pm 8$.

When $x=0$, then $y^2=64 \Rightarrow y= \pm 8$

$$\begin{aligned} x & =8 \text {, then } y^2=64-8^2 \Rightarrow 64-64=0 \\ x & =-8, \text { then } y^2=64-(-8)^2=64-64=0 \\ \therefore \quad R_2 & =\{(0,8),(0,-8),(8,0),(-8,0)\} \end{aligned}$$

We have, $R_3=\{(x,|x|) \mid x$ is real number$\}$

clearly, domain of $R_3=R$

Since, image of any real number under $R_3$ is positive real number or zero.

$$\therefore \quad \text { Range of } R_3=R^{+} \cup\{0\} \text { or }(0, \infty)$$