(6) first choice: the frequency appears higher and wavelength is shorter.
The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.
(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:
[tex]f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz[/tex]
The wavelength is then
[tex]\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m[/tex]
The third choice "0.80m; 431Hz" is correct
