Answer
a) $22.3^\circ$
b) $16.6^\circ$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the aquarium tank is empty, the laser beam will penetrate the glass and then move in the air again before it hits the diffraction grating. Thus, nothing changes here, we have a diffraction grating illuminated by the helium-neon laser.
We know, in a diffraction grating, that the bright fringe angles are given by
$$d\sin\theta_m=m\lambda$$
For the first-order diffraction angle, $m=1$,
$$d\sin\theta_1= \lambda$$
$$\theta_1=\sin^{-1}\left[ \dfrac{\lambda}{d}\right]$$
where $d=1/N$;
$$\theta_1=\sin^{-1}\left[ N\lambda\right]$$
Plugging the known;
$$\theta_1=\sin^{-1}\left[ \dfrac{600}{10^{-3}}\times 633\times 10^{-9}\right]$$
$$\theta_1=\color{red}{\bf 22.3}^\circ$$
$$\color{blue}{\bf [b]}$$
By the same approach,
$$\theta_1'=\sin^{-1}\left[ N\lambda'\right]$$
where $\lambda'$ is the wavelength of the laser beam in the water which is given by $\lambda/n_{water}$
$$\theta_1'=\sin^{-1}\left[ N\dfrac{\lambda}{n_{water}}\right]$$
Plugging the known;
$$\theta_1'=\sin^{-1}\left[ \dfrac{600}{10^{-3}}\cdot\dfrac{ 633\times 10^{-9}}{1.33}\right]$$
$$\theta_1'=\color{red}{\bf 16.6}^\circ$$
