A vibrating string of certain length 'l' under a tension 'T'...

The frequency of the first overtone for an air column inside a tube closed at one end corresponds to the third harmonic, which is given by:

$f = \dfrac{3v}{4L}$

Substituting $v = 340\text{ m/s}$ and $L = 75\text{ cm} = 0.75\text{ m}$:

$f = \dfrac{3 \times 340}{4 \times 0.75} = 340\text{ Hz}$

Since the vibrating string resonates with this mode, the initial frequency of the string is $f_s = 340\text{ Hz}$.

The string produces $4\text{ beats/s}$ with a tuning fork of frequency $n$. Thus, the possible frequencies of the tuning fork are:

$n = f_s \pm 4$

$n = 340 + 4 = 344\text{ Hz}$ or $n = 340 - 4 = 336\text{ Hz}$

The frequency of a vibrating string is directly proportional to the square root of its tension, $f_s \propto \sqrt{T}$. Therefore, increasing the tension increases the frequency of the string ($f_s > 340\text{ Hz}$).

Since increasing the tension reduces the beat frequency, the frequency of the string must move closer to the frequency of the tuning fork. This is only possible if the frequency of the tuning fork is greater than the initial frequency of the string ($n > f_s$).

Hence, $n = 344\text{ Hz}$.