Answer
See the detailed answer below.
Work Step by Step
We are given that
$$\psi(x) =
\begin{cases}
c e^{x/2} & \text{for} \, x \leq 0{\;\rm mm} \\
c e^{-x/2} & \text{for} \, x \geq 0{\;\rm mm}
\end{cases} $$
$$\color{blue}{\bf [a,b,d]}$$
To the wave function and the probability density as functions of $x$, we need to find $c$ first.
We know that
$$ \int_{-\infty}^{\infty} |\psi(x)|^2 \; dx = 1 $$
Plugging from the first formula above.
$$ \int_{-\infty}^{0} |c e^{x/2}|^2 \; dx + \int_{0}^{\infty} |c e^{-x/2}|^2 \; dx = 1 $$
Thus,
$$ \int_{-\infty}^{0} c^2 e^{x} \; dx + \int_{0}^{\infty} c^2 e^{-x} \; dx = 1 $$
$$c^2 \left[ e^{x} \right]_{-\infty}^{0} + c^2 \left[ - e^{-x} \right]_{0}^{\infty} = 1 $$
$$c^2 + c^2 = 1 $$
$$ c^2 = \dfrac{1}{2} $$
$$ \boxed{c = \color{red}{\bf \dfrac{1}{\sqrt{2}}}\;\rm }$$
The two graphs are shown below.
$$\color{blue}{\bf [c]}$$
We know that
$$\text{Prob}(-1.0 \; \text{mm} \leq x \leq 1.0 \; \text{mm}) = \int_{-1.0 \; \text{mm}}^{1.0 \; \text{mm}} |\psi(x)|^2 dx
$$
We can see that the probability density is symmetric around $x=0$, so
$$
\text{Prob}(-1.0 \; \text{mm} \leq x \leq 1.0 \; \text{mm}) = 2 \int_0^{1.0 } c^2 e^{-x } dx $$
$$\text{Prob}(-1.0 \; \text{mm} \leq x \leq 1.0 ) = -2c^2 \left[ e^{-2x/L} \right]_0^{1.0 } $$
$$\text{Prob}(-1.0 \; \text{mm} \leq x \leq 1.0 ) = -2c^2 \left( e^{-1} - 1 \right) $$
$$\text{Prob}(-1.0 \; \text{mm} \leq x \leq 1.0 ) = -2\left[\dfrac{1}{\sqrt2}\right]^2 \left( e^{-1} - 1 \right) $$
$$\text{Prob}(-1.0 \; \text{mm} \leq x \leq 1.0 ) = 0.632=\color{red}{\bf 63.2\%}$$
