Explanation

(a)
$aRb \Leftrightarrow a = 2^{k}.b$ for some integer.
Reflexive $\therefore aRb$ for $k = 0$
Symmetric $aRb \Leftrightarrow a = 2^{k}b$
$\Rightarrow b = 2^{- k}a \Leftrightarrow bRa$
Transitive $aRb \Leftrightarrow a = 2^{k_{1}}b$
$\ \ \ \ \ \ \ \ \ bRc \Leftrightarrow b = 2^{k_{2}}c$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = 2^{k_{1}}.2^{k_{2}}c$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = 2^{k_{1} + k_{2}}c \Leftrightarrow aRc$
$\Rightarrow aRb,\ bRc \Rightarrow aRc$
$\therefore R$ is an equivalent relation.