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 a triangle whose area is given by
$$A=\frac{1}{2}ba$$
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).
Hence,
$$a=\dfrac{2A}{b}=\dfrac{2(1)}{4}=\color{red}{\bf 0.5}\;\rm nm^{-1}$$
$$\color{blue}{\bf [b]}$$
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 $a=0.5$, will be $=\frac{1}{\sqrt2}\approx 0.707$ at $x=0$.
Now what about the point at $x=1$ at which $|\psi(x)|^2=0.25$? it will be the square root as well, $\psi(x)=\frac{1}{2}$
The two possible graphs are 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,
$$\text{Prob}( \text{in } 1\leq x\leq 2 ) = \frac{1}{2}\frac{a}{2}b$$
$$\text{Prob}( \text{in } 1\leq x\leq 2 ) = \frac{1}{2}\cdot \frac{0.5}{2}\cdot (2-1)=\color{red}{\bf 0.125}$$
