What is meant by "spectral moment"?
Assume low-pass signals
throughout.

Since $X(f)$ is usually complex-valued, using the power spectrum $|X(f)|^2$ 
is probably a better idea, especially if you want to take square roots 
etc. afterwards. Thus, $m_k$ is defined as
$$m_k = \int_{-\infty}^\infty f^k |X(f)|^2 \mathrm df.$$
Note in particular that $m_0$ is the power in the signal, and $m_1 = 0$
Now, the Gabor bandwidth $G$ of a signal is given by
$$G = \sqrt{\frac{\displaystyle \int_{-\infty}^\infty f^2 |X(f)|^2 \mathrm df}{\displaystyle\int_{-\infty}^\infty |X(f)|^2 \mathrm df}} = \sqrt{\frac{m_2}{m_0}}.$$
To put this in a slightly different perspective, $|X(f)|^2$ is a nonnegative function,
and the "area under the curve $|X(f)|^2$," viz. $m_0$, is the power in the signal.  Therefore, $|X(f)|^2/m_0$ is effectively a probability density function of 
a zero-mean random variable whose variance is
$$\sigma^2 = \int_{-\infty}^\infty f^2 \frac{|X(f)|^2}{m_0} \mathrm df
= \frac{\displaystyle\int_{-\infty}^\infty f^2 |X(f)|^2 \mathrm df}{\displaystyle\int_{-\infty}^\infty |X(f)|^2 \mathrm df} = G^2$$.

A sinusoid of frequency $G$ Hz has $2G = 2\sqrt{\frac{m_2}{m_0}}$ 
zero crossings per second.  Since Mohammad is reading a legacy book,
it may well be doing all this in radian frequency $\omega$, and thus
if $G$ is the Gabor bandwidth in radians per second, we need to divide
by $2\pi$ giving
$$N_0 = \frac{1}{\pi} \sqrt{\frac{m_2}{m_0}} ~ \text{zero crossings per second.}$$

Turning to extrema, the derivative of $x(t)$ has Fourier transform
$j2\pi f X(f)$ and power spectrum $|2\pi f X(f)|^2$.  Its Gabor
bandwidth is 
$$\begin{align*}G^\prime &amp;= \sqrt{\frac{\displaystyle\int_{-\infty}^\infty f^2 |2\pi f X(f)|^2 \mathrm df}{\displaystyle\int_{-\infty}^\infty |2\pi f X(f)|^2 \mathrm df}}\\
&amp;= \sqrt{\frac{\displaystyle\int_{-\infty}^\infty f^4 |X(f)|^2 \mathrm df}{\displaystyle\int_{-\infty}^\infty f^2 |X(f)|^2 \mathrm df}}\\
&amp;= \sqrt{\frac{m_4}{m_2}}.
\end{align*}$$
Using the same arguments as before (two zero-crossings of the derivative
per period is the same as two extrema per period), radian versus
Hertzian frequency, we get
$$N_e = \frac{1}{\pi} \sqrt{\frac{m_4}{m_2}} ~ \text{extrema per second.}$$