Answer
a) $0\%$
b) $2\%$
c) $\approx 25\%$
Work Step by Step
We know that the probability of detecting a particle in a region of $\delta x$ is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x)=P(x)\delta x =\dfrac{\delta x}{x}$$
Our particle here is a sphere that goes back and forth on a 100-mm track. This means that the center of this sphere, which has a radius of 0.5 mm, is limited to the range of ${\rm 0.50\; mm }\leq x\leq {\rm 99.5\; mm}$. Thus, its total range is 99 mm.
$$\color{blue}{\bf [a]}$$
So, the probability of finding the sphere's center at exactly $x=0.5$ mm is finding it in a region of $\delta x=0$.
Thus,
$$\text{Prob}( \text{in } \delta x=0 \text{ at } x=0.5{ \rm \;mm})=P(x)(0) =\dfrac{0}{x}=\color{red}{\bf 0=0\%}$$
$$\color{blue}{\bf [b]}$$
$$\text{Prob}( \text{in } \delta x \text{ at } {\rm 49\; mm }\leq x\leq {\rm 51\; mm}) =\dfrac{51-49}{99}$$
$$\text{Prob}( \text{in } \delta x \text{ at } {\rm 49\; mm }\leq x\leq {\rm 51\; mm})=\color{red}{\bf 0.0202=2.0\%}$$
$$\color{blue}{\bf [c]}$$
$$\text{Prob}( \text{in } \delta x \text{ at } x \geq{\rm 75\; mm } ) =\dfrac{99.5-75}{99}$$
$$\text{Prob}( \text{in } \delta x \text{ at } {\rm 49\; mm }\leq x\leq {\rm 51\; mm})=\color{red}{\bf 0.247=24.7\%}$$
