Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that
$$\int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1\tag 1$$
We have here a wave function that is shown as a straight line passing through the origin.
The slope of this line is given by ${\rm slope}=\dfrac{c}{4}$
So,
$$\psi(x)=\dfrac{c}{4}x$$
Plug into (1);
$$1=\int_{-4}^{4} \left[\dfrac{c}{4}x\right]^2 dx=\dfrac{c^2}{16} \int_{-4}^{4} x^2 dx$$
$$1 =\dfrac{2c^2}{16} \int_{0}^{4} x^2 dx=\dfrac{c^2}{24} x^3\bigg|_{0}^{4} $$
$$1 = \dfrac{8c^2}{3} $$
Thus,
$$c=\color{red}{\bf\sqrt{\frac{3}{8}}}\;\rm mm^{-1/2}\approx \bf 0.612\;\rm mm^{-1/2}$$
$$\color{blue}{\bf [b]}$$
The graph of the probability density is the square of the $\psi(x)$ graph.
Noting that, at $x=4$ mm, $\psi(x)=\sqrt{3/8}$, hence $|\psi(x)|^2=3/8=0.375$ mm
At $x=2$, mm, $\psi(x)=\sqrt6/16$, hence $|\psi(x)|^2=3/32 $ mm.
$$\color{blue}{\bf [c]}$$
See the figure below.
$$\color{blue}{\bf [d]}$$
We know that
$$\text{Prob}(-2\; \text{mm} \leq x \leq 2\; \text{mm}) = \int_{-2}^{2} |\psi(x)|^2 dx $$
$$\text{Prob}(-2\; \text{mm} \leq x \leq 2\; \text{mm}) = 2\int_{0}^{2} |\psi(x)|^2 dx $$
where $\psi(x)$ in this region is given by $\dfrac{\sqrt{3/8}}{4}x$, so
$$\text{Prob}(-2\; \text{mm} \leq x \leq 2\; \text{mm}) = 2\int_{0}^{2} \frac{3}{128}x^2 dx $$
$$\text{Prob}(-2\; \text{mm} \leq x \leq 2\; \text{mm}) = \frac{3}{64\times 3}\;\;x^3\bigg|_{0}^{2} =\frac{1}{8} $$
$$\text{Prob}(-2\; \text{mm} \leq x \leq 2\; \text{mm})= 0.125=\color{red}{\bf 12.5}\%$$
