(i) Let us assume, to the contrary that $3+\sqrt{3}$ is rational.

Then, $3+\sqrt{3}=\frac{p}{q}$, where $p$ and $q$ are coprime and $q \neq 0$.

$\Rightarrow \sqrt{3}=\frac{p}{q}-3$

$\Rightarrow \sqrt{3}=\frac{p-3 q}{q}$

Since, p and q are are integers.

$\Rightarrow \frac{p-3 q}{q}$ is rational.

So, $\sqrt{3}$ is also rational.

But this contradicts the fact that $\sqrt{3}$ is irrational.

This contradiction has arisen because of our incorrect assumption that $3+\sqrt{3}$ is rational.

Hence, $3+\sqrt{3}$ is irrational.

(ii) Let us assume, to the contrary, that $\sqrt{7}-2$ is rational.

Then, $\sqrt{7}-2=\frac{p}{q}$, where $p$ and $q$ are coprime and $q \neq 0$.

$\Rightarrow \sqrt{7}=\frac{p}{q}+2$

$\Rightarrow \sqrt{7}=\frac{p+2 q}{q}$

Since, p and q are are integers.

$\Rightarrow \frac{p+2 q}{q}$ is rational.

So, $\sqrt{7}$ is also rational.

But this contradicts the fact that $\sqrt{7}$ is irrational.

This contradiction has arisen because of our incorrect assumption that $\sqrt{7}-2$ is rational.

Hence, $\sqrt{7}-2$ is irrational.

(iii) As, $\sqrt[3]{5} \times \sqrt[3]{25}$

$=\sqrt[3]{5 \times 25}$

$=\sqrt[3]{125}$

$=5$, which is an integer

Hence, $\sqrt[3]{5} \times \sqrt[3]{25}$ is rational.

(iv) As, $\sqrt{7} \times \sqrt{343}$

$=\sqrt{7 \times 343}$

$=\sqrt{2401}$

$=49$, which is an integer

Hence, $\sqrt{7} \times \sqrt{343}$ is rational.

(v) As, $\sqrt{\frac{13}{117}}=\sqrt{\frac{1}{9}}=\frac{1}{3}$, which is rational

Hence, $\sqrt{\frac{13}{117}}$ is rational.

(vi) As, $\sqrt{8} \times \sqrt{2}$

$=\sqrt{8 \times 2}$

$=\sqrt{16}$

$=4$, which is an integer

Hence, $\sqrt{8} \times \sqrt{2}$ is rational.