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.
And we have here two identical right triangles whose area is given by
$$A=\frac{1}{2}ba=\frac{1}{2}$$
where $b$ is the base length and $a$ is its height (usually we use h for height but we used a here to match the unknown).
Note that each triangle must have an area of half.
Hence,
$$a=\dfrac{2A}{b}=\dfrac{2\left(\frac{1}{2}\right)}{4}=\color{red}{\bf 0.25}\;\rm fm^{-1}$$
$$\color{blue}{\bf [b]}$$
We know that the graph of $\psi(x)$ is the square root of $|\psi(x)|^2$.
This means that the maximum point on $|\psi(x)|^2$-graph, which is $a=0.25$, will be $=\frac{1}{\sqrt 4}=0.5$ at $x=4,-4$.
Now what about the point at $x=2,-2$ at which $|\psi(x)|^2=\frac{1}{8}$? it will be the square root as well, $\psi(x)=\frac{1}{\sqrt8}\approx 0.354$
The graph is shown below.
$$\color{blue}{\bf [c]}$$
We know that the probability of detecting a neutron is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x)=P(x)\delta x =\text{Area under the curve}$$
So,
$$\text{Prob}( \text{in } |x|\geq 2 ) =\text{Area under the curve}$$
where $|x|\geq 2$ means at $x\geq $ 2 and at $x\leq -2$. The area under the curve here is twice the area of the Trapezoid that starts from point $x=2$ and ends at $x=4$.
So that
$$\text{Prob}( \text{in } |x|\geq 2 ) = 2\times\frac{1}{2} (b_1+b_2)h$$
where $b_1$ and $b_2$ are its two bases and $h$ is its height.
$$\text{Prob}( \text{in } |x|\geq 2 ) = \left(\frac{1}{8}+\frac{1}{4}\right)(4-2)$$
$$\text{Prob}( \text{in } |x|\geq 2 ) = \color{red}{\bf 0.75}$$
