Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The wave function for the ground state $ n = 1 $ of the quantum harmonic oscillator is given by
$$
\psi_1(x) = A_1 e^{-x^2 / 2b^2}\tag 1
$$
The normalization condition requires that the total probability of finding the particle in all space is 1. So
$$
\int_{-\infty}^{\infty} |\psi_1(x)|^2 dx = 1
$$
Substitute $ \psi_1(x) $ from (1);
$$
A_1^2 \int_{-\infty}^{\infty} e^{-x^2 / b^2} dx = 1
$$
To simplify the integral, assume that $ u = \dfrac{x}{b} $, so that $ du = \dfrac{dx}{b} $.
Hence,
$$
A_1^2 b \int_{-\infty}^{\infty} e^{-u^2} du = 1
$$
The integral $ \int_{-\infty}^{\infty} e^{-u^2} du $ is a standard Gaussian integral, and its value is $ \sqrt{\pi} $, so
$$
A_1^2 b\; \sqrt{\pi} = 1
$$
Solve for $ A_1 $;
$$
\boxed{A_1 = \frac{1}{\sqrt{b \sqrt{\pi}}}=\dfrac{1}{(\pi^{1/2}b)^{1/2}}}
$$
Thus, the normalization constant is
$$\boxed{A_1=(\pi^{-1/4})(b^{-1/2})}$$
$$\color{blue}{\bf [b]}$$
The classically forbidden region corresponds to the positions outside the classical turning points, $ |x| > b $.
Because the wave function is symmetrical about $ x = 0 $, the probability of finding the particle in either of the forbidden regions $ x < -b $ or $ x > b $ is twice the probability for just one region. Therefore, the probability is:
$$
{\rm Prob}(x < -b \; {\rm or} \; x > b) =2{\rm Prob}(x < -b \; {\rm or} \; x > b) = 2 \int_b^{\infty} |\psi_1(x)|^2 dx
$$
Substitute $ \psi_1(x) $ from (1);
$$
{\rm Prob}(x < -b \; {\rm or} \; x > b) = 2 \int_b^{\infty} A_1^2 e^{-x^2 / b^2} dx= 2 A_1^2\int_b^{\infty} e^{-x^2 / b^2} dx
$$
Substitute $A_1 $ from the boxed formula above,
$$
\boxed{{\rm Prob}(x < -b \; {\rm or} \; x > b)=\dfrac{ 2}{\sqrt{\pi}\;b} \int_b^{\infty} e^{-x^2 / b^2} dx}
$$
$$\color{blue}{\bf [c]}$$
To find the magnitude, we need to make a change of variables $ u =\dfrac{ x }{ b} $, so $dx=bdu$
Hence, the integral becomes
$$
{\rm Prob}(x < -b \; {\rm or} \; x > b)= \frac{2b}{\sqrt{\pi} \;b} \int_1^{\infty} e^{-u^2} du$$
$$
{\rm Prob}(x < -b \; {\rm or} \; x > b)= \frac{2 }{\sqrt{\pi} } \int_1^{\infty} e^{-u^2} du
$$
The integral $ \int_1^{\infty} e^{-u^2} du $ is computed numerically, and its value is approximately $ 0.139 $.
Thus, the total probability of finding the particle in the forbidden region is:
$$
{\rm Prob}(x < -b \; {\rm or} \; x > b)= \frac{2(0.139 ) }{\sqrt{\pi} } =\bf 0.157
$$
$$
{\rm Prob}(x < -b \; {\rm or} \; x > b) =\color{red}{\bf15.7\%}
$$
