Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The value of the wave function must be the same at $x = 0$ because the wave function must be a continuous function.
Noting that the point $x = 0$ mm is included in both the left interval $-1\; \rm{ mm} \leq x \leq 0\; \rm{ mm}$ and the right interval $0 \;\rm{ mm} \leq x \leq 1 \;\rm{ mm}$),
So, at $x=0$,
$$ \frac{b}{1 +x^2} = c(1 +x)^2 $$
$$ \frac{b}{1 +0^2} = c(1 +0)^2 $$
Thus,
$$ \boxed{b = c }$$
$$\color{blue}{\bf [b]}$$
Since $c=b$, the wave function is given by
$$\psi(x) =
\begin{cases}
\dfrac{b}{(1 + x^2)} & -1 \, \text{mm} \leq x < 0 \, \text{mm} \\\\
b(1 + x)^2 & 0 \, \text{mm} \leq x \leq 1 \, \text{mm}\\\\
0&\rm {elsewhere}
\end{cases}
$$
And Since $c=b$, the probability density function is given by
$$|\psi(x)|^2 =
\begin{cases}
\dfrac{b^2}{(1 + x^2)^2} & -1 \, \text{mm} \leq x < 0 \, \text{mm} \\\\
b^2(1 + x)^4 & 0 \, \text{mm} \leq x \leq 1 \, \text{mm}\\\\
0&\rm {elsewhere}
\end{cases}
$$
See the two graphs below.
$$\color{blue}{\bf [c]}$$
To the right from the origin means from 0 to 1 mm since at $x\gt 1$, $|\psi(x)|^2=0$.
So to find this probability, we need first to find $c$.
We know that
$$\text{Prob}(0\leq x\leq 1)= \int_{-\infty}^{\infty} |\psi(x)|^2 dx =1\tag 1$$
So,
$$\int_{-1}^{0} \frac{c^2}{(1 + x^2)^2} dx + \int_{0}^{1} c^2 (1 + x)^4 dx = 1 \tag 2$$
Solving the first one:
$$
c^2 \int_{-1}^{0} \frac{dx}{(1 + x^2)^2} = c^2 \left[ \frac{x}{2(1 + x^2)} + \frac{\tan^{-1} x}{2} \right]_{-1}^{0} $$
$$c^2 \int_{-1}^{0} \frac{dx}{(1 + x^2)^2} = \left( \frac{1}{4} + \frac{\pi}{8} \right) c^2 = 0.643c^2\tag 3
$$
Solving the second one:
$$
\int_{0}^{1} c^2 (1 + x)^4 dx = c^2 \int_{0}^{1} (x^4 + 4x^3 + 6x^2 + 4x + 1) dx$$
$$ \int_{0}^{1} c^2 (1 + x)^4 dx = c^2 \left[ \frac{x^5}{5} + \frac{4x^4}{4} + \frac{6x^3}{3} + \frac{4x^2}{2} + x \right]_{0}^{1} $$
$$ \int_{0}^{1} c^2 (1 + x)^4 dx = \frac{31}{5} c^2 = 6.20c^2
\tag 4$$
Plug (3) and (4) into (2);
$$ \left( \frac{1}{4} + \frac{\pi}{8} \right) c^2 + \frac{31}{5} c^2 = 1 $$
Thus,
$$c=b \approx \pm \bf 0.382 $$
Hence, from (1)
$$\text{Prob}(0\leq x\leq 1)= \int_{-\infty}^{\infty} |\psi(x)|^2 dx = \int_{0}^{1 } c^2 (1 + x)^4 dx $$
$$\text{Prob}(0\leq x\leq 1)= \int_{0}^{1 } c^2 (1 + x)^4 dx $$
$$\text{Prob}(0\leq x\leq 1) = c^2 \int_{0}^{1 \; } (x^4 + 4x^3 + 6x^2 + 4x + 1) dx $$
$$ \text{Prob}(0\leq x\leq 1) = (0.382)^2 \left[ \frac{x^5}{5} + \frac{4x^4}{4} + \frac{6x^3}{3} + \frac{4x^2}{2} + x \right]_{0}^{1 } $$
$$ \text{Prob}(0\leq x\leq 1) = (0.382)^2 \left[ \frac{1}{5} + 1 + 2 + 2 + 1 \right] $$
$$ \text{Prob}(0\leq x\leq 1) = 0.905 =\color{red}{\bf 90.5\%}$$
