Answer
See a detailed answer below.
Work Step by Step
a) We know, from simple systems, that the period is given by
$$T=2\pi\sqrt{\dfrac{m}{k}}\tag 1$$
Solving for $k$;
$$k=\dfrac{4\pi^2 m}{T^2}\tag 2$$
And since the period time $T$ will change by $dT$ when $m$ changes to $dm$, so taking the first derivative with respect to $m$ for (1). But first, we need to rewrite it as follows:
$$T=2\pi k^{-\frac{1}{2}} (m)^{\frac{1}{2}}$$
where $T$ is a function in $m$
$$\dfrac{d}{dm}(T)=\dfrac{d}{dm}\left[2\pi k^{-\frac{1}{2}} (m)^{\frac{1}{2}} \right] $$
$$\dfrac{dT}{dm}= \color{red}{\bf\not} 2\pi k^{-\frac{1}{2}} \dfrac{1}{\color{red}{\bf\not} 2} (m)^{-\frac{1}{2}} $$
$$\dfrac{dT}{dm}= \pi k^{-\frac{1}{2}} (m)^{-\frac{1}{2}}=\dfrac{\pi}{\sqrt{km}} $$
Thus,
$$ dT = \dfrac{\pi dm}{\sqrt{km}} $$
Plugging from (2);
$$ dT =\dfrac{\pi dm}{\sqrt{\left[\dfrac{4\pi^2 m}{T^2}\right]m}} $$
$$ dT =\dfrac{\color{red}{\bf\not} \pi T dm}{ 2\color{red}{\bf\not} \pi m} $$
$$ \boxed{dT =\dfrac{ T dm}{ 2 m}} $$
Therefore,
$$\boxed{ \Delta T =\dfrac{ T \Delta m}{ 2 m} }$$
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b) The new period is given by
$$T_{\rm new}=T+\Delta T$$
Plugging from the boxed formula above;
$$T_{\rm new}=T+\dfrac{ T \Delta m}{ 2 m} $$
$$T_{\rm new}=T\left[1 +\dfrac{ \Delta m}{ 2 m} \right]$$
Plugging the known, where $\Delta m=0.1\% m=\dfrac{0.1}{100}m$
$$T_{\rm new}=2.0\left[1 +\dfrac{ 10^{-3}\color{red}{\bf\not} m}{ 2 \color{red}{\bf\not} m} \right]$$
$$T_{\rm new}=\color{red}{\bf 2.001}\;\rm s$$
