Two tuning forks A and B give 5 beats per second. Fork A resonates with a closed air column 20 cm long and fork B with an open air column 40.5 cm long. Neglecting end correction, the frequencies of tuning forks $A$ and $B$ are respectively.

The tuning forks produce 5 beats per second, establishing the relationship |$f_A - f_B$| = 5 Hz.

Tuning fork A resonates with a closed air column of length $L_A = 0.20$ m. The fundamental frequency is $f_A = \frac{v}{4L_A} = \frac{v}{0.8}$.

Tuning fork B resonates with an open air column of length $L_B = 0.405$ m. Its fundamental frequency is $f_B = \frac{v}{2L_B} = \frac{v}{0.81}$.

Since $0.8 < 0.81$, it follows that $f_A > f_B$, yielding $f_A - f_B = 5$ Hz. Substituting the expressions:

$\frac{v}{0.8} - \frac{v}{0.81} = 5$
$v\left(\frac{1}{0.8} - \frac{1}{0.81}\right) = 5$
$v\left(\frac{0.01}{0.648}\right) = 5$
$v = \frac{5 \times 0.648}{0.01} = 324$ m/s

The frequencies are then calculated:
$f_A = \frac{324}{0.8} = 405$ Hz
$f_B = \frac{324}{0.81} = 400$ Hz

These values correspond to option C, so the correct choice is $\boxed{\text{C}}$.