Explanation

(a)
Let $x$ and $y$ be two arbitary elements in $A$.
Then, $f(x) = f(y)$
$\Rightarrow \ \ \ \ \ \frac{x - 2}{x - 3} = \frac{y - 2}{y - 3}$
$\Rightarrow xy - 3x - 2y + 6 = xy - 3y - 2x + 6$
$\Rightarrow x = y,\ \forall\ x,\ y \in A$
So, $f$ is an injective mapping.
Again, let $y$ be an orbitary element in $B$, then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(x) = y$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{x - 2}{x - 3} = y$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{3y - 2}{y - 1}$
Clearly, $\forall\ y \in B,\ x = \frac{3y - 2}{y - 1} \in A$, thus for all $y \in B$ there exists $x \in A$ such that
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ f(x) = f\left( \frac{3y - 1}{y - 1} \right) = \frac{\frac{3y - 2}{y - 1} - 2}{\frac{3y - 2}{y - 1} - 3} = y$
Thus, every element in the codomain $B$ has its preimage in $A$, so $f$ is a surjection. Hence, $f:A \rightarrow B$ is bijective.