A convex lens made of glass $\left({\mu }_g=\frac{3}{2}\right)$ has focal length $f$ in air. The image of an object placed in front of it is real, inverted and magnified. Now the whole arrangement is immersed in water $\left({\mu }_w=\frac{4}{3}\right)$ without changing the distance between object and lens, then 

Apply Lens Maker’s Formula, we get 

$\frac{1}{f_{\text{air }}}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f_{\text{water }}}=\left(\frac{3/2}{4/3}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

From these two equations we get, 

$f_{\text{water }}=4f_{\text{air }}=4f$

In air, the image was inverted, real and magnified i.e., the object must be lying between$f \mathrm{and} 2f$. Now the focal length has changed to $4f$. Therefore, the object now lies between pole and focus and so the new image formed will be virtual and magnified. 

Hence, $\left(1\right),\left(3\right)$ are correct.