Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the electrons are uniformly distributed over the interval $(0\leq x\leq 2)$, in cm, with no electrons falling outside, the graph of $|\psi(x)|^2$ must be a rectangle.
And since the area under the graph of $|\psi(x)|^2$ must be 1, the height of this rectangle is given by
$$A=bh$$
where $b$ is the width in the $x$-direction and $h$ is the height in the $y$-direction.
$$h=\dfrac{A}{b}=\dfrac{1}{2}=\bf 0.5\;\rm cm^{-1}$$
$$\color{blue}{\bf [b]}$$
We know that the probability of detecting a particle in a region of $\delta x$ is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x)=P(x)\delta x =|\psi(x)|^2\delta x$$
Plug the known;
$$\text{Prob}( \text{in } \delta x\text{ at } 0.8{\;\rm cm})= |\psi(0.8)|^2(0.81-0.79)$$
where $\delta x=0.81-0.79=0.02$ cm,
$$\text{Prob}( \text{in } \delta x\text{ at } 0.8{\;\rm cm})= (0.5)(0.02)=0.01=\color{red}{\bf 1.0}\%$$
$$\color{blue}{\bf [c]}$$
Recalling that
$$\text{Prob}( \text{in } \delta x \text{ at } x)=\dfrac{N}{N_{tot}}$$
where $N$ is the number of detected electrons while $N_{tot}$ is the total number of electron.
Thus,
$$N=N_{tot}\cdot \text{Prob}( \text{in } \delta x\text{ at } 0.8{\;\rm cm})$$
Plug the known
$$N=(10^6)(0.01)=\color{red}{\bf 10^4}\;\rm electron$$
$$\color{blue}{\bf [d]}$$
The probability density is at $x=0.8$ cm is given by
$$P(x)=|\psi(x)|^2$$
$$P(0.8)=|\psi(0.8)|^2=\color{red}{\bf 0.5}\;\rm cm^{-1}$$
