Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the area under the probability density curve is 1.
So we have to draw the $|\psi(x)|^2$ graph to find $c$.
We know that the $|\psi(x)|^2$ graph is the square of $|\psi(x)|$.
This means that the maximum point on $|\psi(x)|$-graph, which is $c$, will be $c^2$, and the minimum point of $-\frac{1}{2}c$ will be $\frac{1}{4}c^2$
The graph is shown below.
Now we can find $c$; where the area under this curve consists of two rectangles.
$$A=(4)(0.25c^2)+(0.75c^2)(2)=1$$
where $(4)(0.25c^2)$ is the area of the first rectangle that its base from $x=-2$ to $x=-2$ and its height is $0.25c^2$, and $(0.75c^2)(1)$ is the other (upper) rectangle.
Hence,
$$c=\dfrac{\sqrt{10}}{5}=\color{red}{\bf 0.632}\;\rm nm^{-1/2}$$
$$\color{blue}{\bf [b]}$$
The graph is shown below.
$$\color{blue}{\bf [c]}$$
We know that the probability of detecting an electron is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x)=P(x)\delta x =\text{Area under the curve}$$
So,
$$\text{Prob}( \text{in } -1\leq x\leq 1 ) =\text{Area under the $|\psi(x)|^2$ curve}$$
So that
$$\text{Prob}( \text{in } -1\leq x\leq 1 ) =(2)(c^2)=(2)\left( \frac{2}{5}\right)$$
$$\text{Prob}( \text{in } -1\leq x\leq 1 ) = \color{red}{\bf 0.8}$$
