Answer
a) $0.271\%$
b) $31.8\%$
Work Step by Step
The given wave function for the electron is normalized, and we know that the probability of finding the electron in a given region is the integral of the square of the wave function $ |\psi(x)|^2 $ over that region.
$$ \psi(x) = \begin{cases}
0 & x < 0 \; \text{nm} \\
(1.414 \; \text{nm}^{-1/2}) e^{-x/(1.0 \; \text{nm})} & x \geq 0 \; \text{nm}
\end{cases} $$
$$\color{blue}{\bf [a]}$$
The probability of finding the electron in a $\delta x=\rm 0.010 \;nm$- wide region at $ x = 1.0 \; \text{nm} $ is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x=1.0{\;\rm nm}) = |\psi(1.0)|^2 (0.010)$$
Plugging $\psi(x)$ from the formula above for $x\geq 0$
$$\text{Prob}( \text{in } \delta x \text{ at } x=1.0{\;\rm nm}) = (0.010)\left|(1.414 ) e^{-1.0/(1.0)}\right|^2 $$
$$\text{Prob}( \text{in } \delta x \text{ at } x=1.0{\;\rm nm}) = \bf 0.00271=\color{red}{\bf 0.271\%}$$
$$\color{blue}{\bf [b]}$$
The probability of finding the electron in the interval $ 0.50 \; \text{nm} \leq x \leq 1.50 \; \text{nm} $ is given by
$$ P(0.50 \leq x \leq 1.50) = \int_{0.50}^{1.50} |\psi(x)|^2 \; dx\tag 1 $$
where the wave function $ \psi(x) $ for $ x \geq 0 $ is given by:
$$ \psi(x) = (1.414 \; \text{nm}^{-1/2}) e^{-x/(1.0 \; \text{nm})} $$
Thus, the square of the wave function is:
$$ |\psi(x)|^2 = (1.414)^2 e^{-2x/(1.0 \; \text{nm})} = 2.0 \; e^{-2x} $$
Plug into (1)
$$ P(0.50 \leq x \leq 1.50) =2.0 \int_{0.50}^{1.50} e^{-2x} \; dx $$
$$ P(0.50 \leq x \leq 1.50) =2.0\left[-\frac{1}{2} e^{-2x} \right]_{0.50}^{1.50} $$
$$ P(0.50 \leq x \leq 1.50) = -e^{-2x} \bigg|_{0.50}^{1.50} $$
$$ P(0.50 \leq x \leq 1.50) = \left( -e^{-2(1.50)} \right) - \left( -e^{-2(0.50)} \right) $$
$$P(0.50 \leq x \leq 1.50) =\bf0.318=\color{red}{\bf 31.8\%}$$
