Answer
a.) $A=\sqrt{\frac{\lambda}{\pi}}$
b.) $\langle x\rangle=a$, $\langle x^2\rangle=a^2 + \frac{1}{2\lambda}$
c.) The plot should be a gaussian distrubtion. See below.
Work Step by Step
a.) As instructed, we are to use Eq. 1.16 in the book which is given by the following,
$$1=\int_{-\infty}^{\infty} \rho(x) dx.$$
Substituting the given distribution $\rho(x)=A e^{-\lambda(x-a))^2}$ leads to the following equation,
$$1=\int_{-\infty}^{\infty} A e^{-\lambda(x-a))^2} dx.$$
The integral above can be evaluated by introducing the following variables, $u=x-a, du=dx, u: -\infty \to \infty$. This leads to,
$$1=A \int_{-\infty}^{\infty} e^{-\lambda u^2} dx.$$
The integral above is just the known Gaussian integral, that is, $\int_{-\infty}^{\infty} e^{-\lambda u^2} dx=\sqrt{\pi/\lambda}.$ Hence,
$$1=A\sqrt{\frac{\lambda}{\pi}}$$
$$A=\sqrt{\frac{\lambda}{\pi}}.$$
b.) Calculate first for $\langle x \rangle$.
$$\langle x \rangle=A\int_{\infty}^{\infty}x e^{-\lambda(x-a)^2 dx}$$
Use the same variables as in part a. This leads to
$$\langle x \rangle=A\int_{\infty}^{\infty}(u+a) e^{-\lambda u^2 dx}=A\left[\int_{\infty}^{\infty}u e^{-\lambda u^2 dx}+a\int_{\infty}^{\infty} e^{-\lambda u^2 dx}\right]$$
$$=A\left(0+a\sqrt{\frac{\pi}{\lambda}}\right) = a$$
Next is $\langle x^2 \rangle$.
$$\langle x \rangle=A\int_{\infty}^{\infty}x^2 e^{-\lambda(x-a)^2 dx}$$
$$=A\left[\int_{\infty}^{\infty}u^2 e^{-\lambda u^2 dx}+2a\int_{\infty}^{\infty} e^{-\lambda u^2 dx}+\int_{\infty}^{\infty}a^2 e^{-\lambda u^2 dx}\right]$$
$$=A\left[\frac{1}{2\lambda}+0+a^2\frac{\pi}{\lambda}\right]$$
$$=a^2+\frac{1}{2\lambda}$$
Finally,
$$\sigma^2=\langle x \rangle-\langle x^2 \rangle=a^2 + \frac{1}{2\lambda}-a^2=\frac{1}{2\lambda}$$
c.) See plot. As can be seen, it is a gaussian distribution whose peak is located at (a,A).
