Answer
a) $10\;\rm s$
b) $3.83\;\rm m/s^2$
c) $6.4\%$
Work Step by Step
a) To find the race time, we need to find the distance during the acceleration.
$$x_1=x_0+v_{0x}t_1+\frac{1}{2}a_xt_1^2=0+0+\left(\frac{1}{2}\cdot 3.6\cdot\left[\dfrac{10}{3}\right]^2 \right)$$
$$x_1=20\;\rm m\tag 1$$
Thus, the rest distance will be at a constant speed which is given by
$$v_{x1}=v_{0x}+a_xt_1=0+3.6\cdot \dfrac{10}{3} $$
$$v_{x1}=12\;\rm m/s\tag 2$$
This is the speed the sprinter completes the race at.
$$\Delta x=100-x_1=v_{x1}t_2$$
Solving for $t_2$;
$$t_2=\dfrac{100-x_1}{v_{x1}}$$
Thus the total time is given by
$$t=t_1+t_2=\dfrac{10}{3}+\dfrac{100-x_1}{v_{x1}}$$
Plugging from (1) and (2);
$$t= \dfrac{10}{3}+\dfrac{100-20}{12}$$
$$t=\color{red}{\bf10}\;\rm s$$
b) The whole distance is given by
$$x=x_1+v_{x1}t_2$$
$$100=x_1+12t_2$$
where $x_1=\frac{1}{2}a_xt_1^2$
$$100=\frac{1}{2}a_xt_1^2+12t_2\tag 3$$
$$v_{x1}=v_0+a_xt_1$$
$$12=a_xt_1\tag{Plug into (3)}$$
$$100=\frac{1}{2}t_1 \;\overbrace{ (a_xt_1) }^{12} +12t_2 $$
$$100=6t_1+12t_2$$
Noting that $t_1+t_2=9.9$, thus, $t_2=9.9-t_1$
$$100=6t_1+12(9.9-t_1)$$
Hence,
$$t_1=3.13\;\rm s$$
Thus, $$a_x=\dfrac{12}{3.13}$$
$$a_x=\color{red}{\bf3.83}\;\rm m/s^2$$
c) Decreasing his time by 1$\%$ means he will run the 100-m race in
$$t=t_{original}-0.01t_{original}=10-(0.01\cdot 10)=\bf 9.9\;\rm s$$
which is the same time for the second case when his acceleration is 3.83 m/s$^2$.
Thus, the percentage increase in his acceleration is given by
$${ a_x\%}_{\rm increase}=\dfrac{a_{x2}-a_{x1}}{a_{x1}}\times100\%$$
$${ a_x\%}_{\rm increase}=\dfrac{3.83-3.6 }{3.6}\times100\%$$
$${ a_x\%}_{\rm increase}=\color{red}{\bf 6.4}\%$$
