$\left( 1 \right)$ ${}^{1}/{}_{\sqrt{2}}$
$(2)7\sqrt{5}$
$(3)6+\sqrt{2}$
$\left( 1 \right)$ Let us assume, ${}^{1}/{}_{\sqrt{2}}$ is rational.
Then, we can find co-prime x and y$(y\ne 0)$ such that ${}^{1}/{}_{\sqrt{2}}={}^{x}/{}_{y}$
Rearranging,we get,
$\sqrt{2}=\frac{y}{x}$
Since,x and y are integers ,Thus $\sqrt{2}$ is a rational number,
which contradicts the fact that $\sqrt{2}$ is irrational.
Hence, we can conclude that ${}^{1}/{}_{\sqrt{2}}$ is irrational number.
$\left( 2 \right)$ Let us assume $7\sqrt{5}$ is a rational number.
Then we can find co-prime a and b $\left( b\ne 0 \right)$ such that $7\sqrt{5}={}^{x}/{}_{y}$
Rearranging, we get,
$\sqrt{5}=\frac{x}{y}$ Since, x and y are integers, thus, $\sqrt{5}$ is a rational number, which contradicts the fact that $\sqrt{5}$ is irrational.
Hence, we can conclude that $7\sqrt{5}$ is irrational number
(3)$6+\sqrt{2}$
Let us assume $6+\sqrt{2}$ is a rational number.
Then we can find co-primes x and y ($y\ne 0$ ) such that $6+\sqrt{2}={}^{x}/{}_{y}$
Rearranging, we get,
$\sqrt{2}={}^{x}/{}_{y}-6$
Since, x and y are integers, thus ${}^{x}/{}_{y}-6$ is a rational number and therefore, $\sqrt{2}$ is rational. This contradicts the fact that $\sqrt{2}$ is an irrational number.
Hence, we can conclude that $6+\sqrt{2}$ is irrational number.