(3)

As, intensity $I\propto$ width of slit W

Also, intensity $I\propto$ square of amplitude A

$\therefore$    $\frac{I_1}{I_2}=\frac{W_1}{W_2}=\frac{A_1^2}{A_2^2}$

But  $\frac{W_1}{W_2}=\frac{1}{25}$  (given)     $\therefore$  $\frac{A_1^2}{A_2^2}=\frac{1}{25} \text{or} \frac{A_1}{A_2}=\sqrt{\frac{1}{25}}=\frac{1}{5}$

$\therefore$   $\frac{I_{\text{max}}}{I_{\text{min}}}=\frac{{(A_1+A_2)}^2}{{(A_1-A_2)}^2}=\frac{{(\frac{A_1}{A_2}+1)}^2}{{(\frac{A_1}{A_2}-1)}^2}$

          $=\frac{{(\frac{1}{5}+1)}^2}{{(\frac{1}{5}-1)}^2}=\frac{{\left(\frac{6}{5}\right)}^2}{{(-\frac{4}{5})}^2}=\frac{36}{16}=\frac{9}{4}$

(3)

Intensity at any point on the screen is

   $I=4I_0{\cos }^2\frac{{\displaystyle \varphi }}{{\displaystyle 2}}$

where $I_0$ is the intensity of either wave and $\varphi$ is the phase difference between two waves.

Phase difference, $\varphi =\frac{2\pi }{\lambda }\times \text{Path difference}$

When path difference is $\lambda ,$ then

     $\varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$

$\therefore$    $I=4I_0{\cos }^2\left(\frac{2\pi }{2}\right)=4I_0{\cos }^2\left(\pi \right)=4I_0=K$                            ...(i)

When path difference is $\frac{\lambda }{4}$, then

$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$     $\therefore$  $I=4I_0{\cos }^2\left(\frac{\pi }{4}\right)=2I_0=\frac{K}{2} \text{[Using (i)]}$