(a) We first find ${R^{ – 1}},$ we have 

${R^{ – 1}} = \{ (5,\,4);\,(4,\,1);\,(6,\,4);\,(6,\,7);\,(7,\,3)\} $.

We now obtain the elements of ${R^{ – 1}}oR$ we first pick the element of

$R$ and then of ${R^{ – 1}}$. Since $(4,\,5) \in R$ and $(5,\,4) \in {R^{ – 1}}$, we have $(4,\,4) \in {R^{ – 1}}oR$

Similarly, $(1,\,4) \in R,\,(4,\,1) \in {R^{ – 1}} \Rightarrow \,(1,\,1) \in {R^{ – 1}}oR$

$(4,\,6) \in R,\,(6,\,4) \in {R^{ – 1}} \Rightarrow \,(4,\,4) \in {R^{ – 1}}oR,$

$(4,\,6) \in R,\,(6,\,7) \in {R^{ – 1}} \Rightarrow \,(4,\,7) \in {R^{ – 1}}oR$

$(7,\,6) \in R,\,(6,\,4) \in {R^{ – 1}} \Rightarrow \,(7,\,4) \in {R^{ – 1}}oR,$

$(7,\,6) \in R,\,(6,\,7) \in {R^{ – 1}} \Rightarrow \,(7,\,7) \in {R^{ – 1}}oR$

$(3,\,7) \in R,\,(7,\,3) \in {R^{ – 1}} \Rightarrow \,(3,\,3) \in {R^{ – 1}}oR,$

Hence,${R^{ – 1}}oR = \{(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)\}.$