Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the wave speed in the string is given by
$$v=\sqrt{\dfrac{T_s}{\mu}}$$
where $\mu=m/L$, and $v=\lambda f$
Hence,
$$f=\dfrac{1}{\lambda}\sqrt{\dfrac{T_s}{\mu }}$$
$$f=\dfrac{1}{\lambda\;\sqrt{\mu}}\sqrt{T_s}\tag 1$$
And when the tension increases by $\Delta T_s$ without changing the length, so $\mu$ is constant, and $\lambda$ is constant as well.
Taking the derivative of both sides with respect to $dT_s$
$$\dfrac{df}{dT_s} =\dfrac{1}{\lambda\;\sqrt{\mu}}\dfrac{d}{dT_s} T_s^{\frac{1}{2}}$$
$$\dfrac{df}{dT_s} =\dfrac{1}{2\lambda\;\sqrt{\mu}} T_s^{\frac{-1}{2}}$$
$$\dfrac{df}{dT_s} =\dfrac{1}{2\lambda\;\sqrt{\mu T_s}} $$
$$\dfrac{df}{dT_s} =\dfrac{1}{2\lambda\;\sqrt{\mu T_s}}\cdot\dfrac{\sqrt{T_s}}{\sqrt{T_s}} $$
$$\dfrac{df}{dT_s} =\dfrac{1}{2T_s\lambda}\sqrt{\dfrac{T}{\mu}} $$
$$\dfrac{df}{dT_s} =\dfrac{1}{2T_s\lambda}v= \dfrac{1}{2T_s\lambda}\lambda f$$
$$\dfrac{df}{dT_s} = \dfrac{f}{2T_s} $$
And hence,
$$\dfrac{df}{f} = \dfrac{dT_s}{2T_s} $$
Therefore,
$$\boxed{\dfrac{\Delta f}{f} = \dfrac{\Delta T_s}{2T_s} }$$
$$\color{blue}{\bf [b]}$$
We know that $\Delta f=5\;\rm Hz$ or beats per second so we can easily find the change in tension by plugging the known into the boxed formula above.
$$\dfrac{\Delta T_s}{T_s}=\dfrac{2 \Delta f}{f}=\dfrac{2(5)}{(500)}$$
$$\dfrac{\Delta T_s}{T_s}=\dfrac{1}{50}=0.02$$
Thus, the percentage increase in the tension is then
$$\dfrac{\Delta T_s}{T_s}=\color{red}{\bf2.0}\bf \%$$
