The equation of a line parallel to $x$-axis is $y=k.$

Solving this equation with $y=\sqrt{x},$ we get $x=k^2.$

Thus, the line $y=k$ intersect with $y=\sqrt{x}$ at $\left(k^2, k\right).$

It is given that the line $y=k$ intersects the curve $y=\sqrt{x}$ at an angle

or $n/4.$This means that the slope of the tangent to $y=\sqrt{x}$ at

$\left({\left.k\right|}^2, k\right)$ is $\tan (\pm \pi /4)=\pm 1.$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(k^2, k\right)}=\pm 1\Rightarrow {\left(\frac{1}{2\sqrt{x}}\right)}_{\left(k^2, k\right)}=\pm 1\Rightarrow k=\pm \frac{1}{2}$

Hence, the required lines are $y=\pm \frac{1}{2}$