Solution:

(i) $\sqrt{2}, 1 / \sqrt{2}, 1 / 2 \sqrt{2}, 1 / 4 \sqrt{2}, \ldots$ is $1 / 512 \sqrt{2} ?$
Using the formula,
$\begin{array}{l}
T_{n}=a r^{n-1} \\
a=\sqrt{2} \\
r=t_{2} / t_{1}=(1 / \sqrt{2}) /(\sqrt{2}) \\
=1 / 2 \\
T_{n}=1 / 512 \sqrt{2} \\
n=? \\
T_{n}=a r^{n-1} \\
1 / 512 \sqrt{2}=(\sqrt{2})(1 / 2)^{n-1} \\
1 / 512 \sqrt{2} \times \sqrt{2}=(1 / 2)^{n-1} \\
1 / 512 \times 2=(1 / 2)^{n-1} \\
1 / 1024=(1 / 2)^{n-1} \\
(1 / 2)^{10}=(1 / 2)^{n-1} \\
10=n-1 \\
n=10+1 \\
=11
\end{array}$
Therefore, $11^{\text {th }}$ term of the G.P is $1 / 512 \sqrt{2}$

(ii) $2,2 \sqrt{2}, 4, \ldots$ is 128 ?
Using the formula,
$\begin{array}{l}
\mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \\
\mathrm{a}=2 \\
\mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(2 \sqrt{2 / 2)} \\
=\sqrt{2} \\
\mathrm{~T}_{\mathrm{n}}=128 \\
\mathrm{n}=? \\
\mathrm{~T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \\
128=2(\sqrt{2})^{\mathrm{n}-1}
\end{array}$
$128 / 2=(\sqrt{2})^{n-1}$ $64=(\sqrt{2})^{n-1}$ $2^{6}=(\sqrt{2})^{n-1}$ $12=n-1$ $n=12+1$ $=13$
Therefore, $13^{\text {th }}$ term of the G.P is 128