Solution:
(i) One-One but not Onto
$\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ be a mapping given by $\mathrm{f}(\mathrm{x})=\mathrm{x} 2$
For one-one
$\begin{array}{l}
f(x)=f(y) \\
x_{2}=y z \\
x=y(x \text { cannot be } y \text { as } y \in N)
\end{array}$
Therefore, $f(x)=x_{2}$ For each element in $\operatorname{Set} A$, it has unique image in $\operatorname{Set} B$ For onto
Assume $f(x)=y$, for $y \in N$
$\mathrm{x} 2=\mathrm{y}$
$\mathrm{x}=\mathrm{y} 1 / 2$
If $\mathrm{y}=2, \mathrm{x}=1.414$, which is not a natural number.
As a result, not onto.
Therefore $f(x)=x_{2}$ is one-one but not onto
(ii) One-one and onto
$\mathrm{f}: \mathrm{R}-\{5\} \rightarrow \mathrm{R}-\{1\}$ given by $\mathrm{f}(\mathrm{x})=\frac{x-7}{x-5}$
For one-one
$\begin{array}{l}
\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{y}) \\
\frac{x-7}{x-5}=\frac{y-7}{y-5} \\
\Rightarrow \mathrm{xy}-5 \mathrm{x}-7 \mathrm{y}+35=\mathrm{xy}-7 \mathrm{x}-5 \mathrm{y}+35 \\
\Rightarrow 2 \mathrm{x}=2 \mathrm{y} \\
\Rightarrow \mathrm{x}=\mathrm{y}
\end{array}$
Therefore, $f(x)=\frac{x-7}{x-5}$ is one-one
For onto
Assume $\mathrm{f}(\mathrm{x})=\mathrm{y}$
$\begin{array}{l}
\Rightarrow \frac{x-7}{x-5}=y \\
\Rightarrow x-7=y(x-5) \\
\Rightarrow x=\frac{7-5 y}{1-y}
\end{array}$
We can clearly see, $x=\frac{7-5 y}{1-y}$ is a real number for all $y \neq 1$
Also, $\frac{7-5 y}{1-y} \neq 5$, because then we get $7=5$, which is not possible.
Hence, for element $y$ in $B$ has its pre image $x$ in A given by $x=\frac{7-5 y}{1-y}$