Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that
$$\int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1$$
1. For $ 0 \leq x \leq 0.75 \;\text{nm} $,
$$ \psi(x) =\dfrac{c}{0.75}x= \dfrac{4c}{3}x $$
2. For $ 0.75 \;\text{nm} \leq x \leq 1.0 \;\text{nm} $,
$$ \psi(x) =\dfrac{c}{0.25}(1.0 - x)= 4c(1.0 - x) $$
Since $\psi(x) = 0 $ outside the interval $[0, 1] $, we only need to integrate between $x = 0 $ and $x = 1.0 \; \rm{nm} $. We split the integration into two parts for each section of the wave function.
$$1= \int_0^{0.75} \left( \frac{4c}{3} x \right)^2 dx + \int_{0.75}^{1.0} \left( 4c(1 - x) \right)^2 dx $$
$$1= \frac{16}{9} c^2 \left[ \frac{x^3}{3} \right]_0^{0.75} + 16 c^2 \left[ x - x^2 + \frac{x^3}{3} \right]_{0.75}^{1.0}$$
$$1= \frac{16}{9} \left[ \dfrac{9}{64} \right] c^2+ 16 \left[ \dfrac{1}{192} \right]c^2 $$
Thus,
$$c=\pm \sqrt{3}$$
and we can dismiss the negative root since $c$ is in the positive $y$-direction.
Therefore,
$$c=\color{red}{\bf \sqrt{3}}\;\rm nm^{-1/2}$$
$$\color{blue}{\bf [b]}$$
The graph of the probability density is the square of $\psi(x)$ graph.
Noting that, at $x=0.75$ nm, $\psi(x)=\sqrt3$, hence $|\psi(x)|^2=3$.
At $x=0.5$, nm, $\psi(x)=2\sqrt3/3$, hence $|\psi(x)|^2=4/3 $.
$$\color{blue}{\bf [c]}$$
See the figure below.
$$\color{blue}{\bf [d]}$$
We know that
$$\text{Prob}(0.0 \; \text{nm} \leq x \leq 0.25 \; \text{nm}) = \int_{0.0}^{0.25} |\psi(x)|^2 dx $$
where $\psi(x)$ in this region is given by $\dfrac{4\sqrt3}{3}x$, so
$$\text{Prob}(0.0 \; \text{nm} \leq x \leq 0.25 \; \text{nm}) = \frac{16}{3} \int_{0.0}^{0.25} x^2 dx $$
$$\text{Prob}(0.0 \; \text{nm} \leq x \leq 0.25 \; \text{nm}) = \frac{16}{3} \left[ \frac{x^3}{3} \right]_{0.0}^{0.25} = \frac{1}{36} $$
$$\text{Prob}(0.0 \; \text{nm} \leq x \leq 0.25 \; \text{nm}) = 0.028=\color{red}{\bf 2.8}\%$$
