Solve it in English

If $\frac{1}{\sqrt{2}}$ is rational, then it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Since $a$ and $b$ are integers and $a \neq 0$, $\frac{b}{a}$ is a rational number.

This implies that $\sqrt{2}$ is a rational number. However, it is a known fact that $\sqrt{2}$ is irrational. This contradicts our initial assumption.

Therefore, our assumption that $\frac{1}{\sqrt{2}}$ is rational must be false.
Hence, $\boxed{\text{}\frac{1}{\sqrt{2}}\text{ is irrational.}}$

If $7\sqrt{5}$ is rational, then it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Since $a$ and $b$ are integers and $7b \neq 0$, $\frac{a}{7b}$ is a rational number.

This implies that $\sqrt{5}$ is a rational number. However, it is a known fact that $\sqrt{5}$ is irrational. This contradicts our initial assumption.

Therefore, our assumption that $7\sqrt{5}$ is rational must be false.
Hence, $\boxed{\text{}7\sqrt{5}\text{ is irrational.}}$

If $6+\sqrt{2}$ is rational, then it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.

Since $a$ and $b$ are integers and $b \neq 0$, $(a - 6b)$ is an integer and $b$ is an integer. Therefore, $\frac{a - 6b}{b}$ is a rational number.

This implies that $\sqrt{2}$ is a rational number. However, it is a known fact that $\sqrt{2}$ is irrational. This contradicts our initial assumption.

Therefore, our assumption that $6+\sqrt{2}$ is rational must be false.
Hence, $\boxed{\text{}6+\sqrt{2}\text{ is irrational.}}$