#### Answer

(a) The original frequency of the untuned string was $290.0~Hz$.
(b) The tension increased by 1.4%

#### Work Step by Step

(a) We can hear beats at a beat frequency that is equal to the difference of the two frequencies of the strings. Since the beat frequency was $3.0~Hz$, the difference in the frequencies of the strings must be $3.0~Hz$. Since the untuned string was at a lower frequency, the original frequency of the untuned string was $290.0~Hz$.
(b) After increasing the tension, the beat frequency was $1.0~Hz$. We can assume that the frequency changed from $290.0~Hz$ up to $292.0~Hz$. We can find the change in tension in the string.
We can write an expression for the original tension $F_1$:
$\sqrt{\frac{F_1}{\mu}} = v_1$
$\sqrt{\frac{F_1}{\mu}} = \lambda~f_1$
$\frac{F_1}{\mu} = (\lambda~f_1)^2$
$F_1 = (\mu)(\lambda~f_1)^2$
We can write an expression for the new tension $F_2$:
$\sqrt{\frac{F_2}{\mu}} = v_2$
$\sqrt{\frac{F_2}{\mu}} = \lambda~f_2$
$\frac{F_2}{\mu} = (\lambda~f_2)^2$
$F_2 = (\mu)(\lambda~f_2)^2$
We can divide the second equation by the first equation:
$\frac{F_2}{F_1} = \frac{(\mu)(\lambda~f_2)^2}{(\mu)(\lambda~f_1)^2}$
$F_2 = \frac{f_2^2}{f_1^2}~F_1$
$F_2 = \frac{(292.0~Hz)^2}{(290.0~Hz)^2}~F_1$
$F_2 = 1.014~F_1$
The tension increased by 1.4%