Given, refractive index of glass, ${\mathrm{\mu }}_{\mathrm{g}}$ = 1.5 

Refractive index is given by, 

             $\mathrm{\mu } = \frac{\mathrm{speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{vacuum}}{\mathrm{speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{the} \mathrm{medium}}$

$\therefore$  Speed of light in glass$= \frac{\mathrm{speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{vaccum}}{{\mathrm{\mu }}_{\mathrm{g}}}$
                                      $$\begin{gathered} = \frac{3.0 \times {10}^6}{1.5} \\ = 2.0 \times {10}^6 {\mathrm{ms}}^{-1}. \end{gathered}$$

Given, 
Wavelength of monochromatic light, $\mathrm{\lambda } = 589 \mathrm{nm}$$$\begin{gathered} \mathrm{Speed} \mathrm{of} \mathrm{light}, \mathrm{c} = 3 \times {10}^8 \mathrm{m}/\mathrm{s} \\ \mathrm{Refractive} \mathrm{index}, \mathrm{\mu } = 1.33 \end{gathered}$$

(a) For reflected light,
Wavelength is going to remain the same for incident and reflected light. 
Wavelength, $\mathrm{\lambda } = 589 \mathrm{nm} = 589 \times {10}^{-9}\mathrm{m}$
$$\begin{gathered} \mathrm{Frequency} \mathrm{of} \mathrm{light}, \mathrm{v} = \frac{\mathrm{c}}{\mathrm{\lambda }} \\ = \frac{3 \times {10}^8}{589 \times {10}^{-9}} \\ = 5.09 \times {10}^{14}\mathrm{Hz}. \end{gathered}$$
Speed of lightwill remain invariant in two media.
Therefore, $\mathrm{v} = \mathrm{c} = 3 \times {10}^8 \mathrm{m}/\mathrm{s}$ 

(b) For refracted light, 
Refractive index of water w.r.t. air,   $\boxed{{}_{\mathrm{a}}\mathrm{\mu }_{\mathrm{\omega }} = \frac{{\mathrm{v}}_{\mathrm{a}}}{{\mathrm{v}}_{\mathrm{\omega }}}}$ 

i.e.,                                                 ${}_{\mathrm{a}}\mathrm{\mu }_{\mathrm{\omega }} = \frac{{\mathrm{v}}_{\mathrm{a}}{\mathrm{\lambda }}_{\mathrm{a}}}{{\mathrm{v}}_{\mathrm{\omega }}{\mathrm{\lambda }}_{\mathrm{\omega }}}$
$\therefore$                 ${\mathrm{v}}_{\mathrm{a}} = {\mathrm{v}}_{\mathrm{\omega }}$
So, wavelength of light 

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{\omega }} = \frac{{\mathrm{\lambda }}_{\mathrm{a}}}{{}_{\mathrm{a}}\mathrm{\mu }_{\mathrm{\omega }}} = \frac{589 \times {10}^{-9}}{1.33} \\ = 4.42 \times {10}^{-7}\mathrm{m} \end{gathered}$$
                          
As frequency remains unaffected on entering another medium, therefore, 
           ${\nu }_{\mathrm{\omega }} = {\nu }_{\mathrm{a}} = 5.09 \times {10}^{14} \mathrm{Hz}$ 

Speed,
              $$\begin{gathered} \mathrm{v}\text{'} = \frac{v_{\mathrm{a}}}{{\mathrm{\mu }}_{\mathrm{\omega }}} = \frac{3 \times {10}^8}{1.33} \\ = 2.25 \times {10}^8 \mathrm{m}/\mathrm{s}. \end{gathered}$$  

i.e., speed of the light has decresed when it entered the second medium.