Explanation

(a) For string fixed at both the ends, resonant frequency are given by
$v = \frac{nv}{2L}$
Where symbols have their meaning. It is given that 315 Hz and 420 Hz are two consecutive resonant frequency, let these nth and (n+1)th harmonics.
$315 = \frac{nv}{2L}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$420 = \frac{(n + 1)v}{2L}\ \ \ \ \ \ \ \ldots(ii)$
⟹Eq.(i)÷Eq.(ii)
$\Longrightarrow \frac{315}{450} = \frac{n}{n + 1} \Longrightarrow n = 3$
From Eq. (i), lowest resonant frequency
$v_{0} = \frac{v}{2L} = \frac{315}{3} = 105\ Hz$