$\varphi _{X}(w)=\sum ^{\infty}_{k=0}(-1)^k(wb)^{2k}$,and the mth moment of X is $E[X^m]=(-j)^m\frac{d^m \varphi _{X}(w)}{dw^m}=(-j)^m\sum_k (-1)^k \times 2k \times(2k-1)\times ... \times(2k-(m-1))\times b^{2k}w^{2k-m}$
From the formula we derive,we found the $2k=m$,and the solution said so $E[X^m]=(-j)^m \times m! \times b^m$,i know where are $(-j)^m$ and $m!$ and $b^m$ from and the reason that $w^{2k-m}$ is dissappeared,but where is the $\sum_k(-1)^k $ going?
Your formula to compute the $m^{th}$ moment is not correct. I assume that $j$ is defined such that $j^2 = -1$ (which is not the standard notation by the way), that $b$ is some positive constant, and that $\phi_X(w) = E[e^{jwX}]$ is well defined for $|w|<\frac{1}{b}$.
To answer your question, the correct formula is: $$ E[X^m] = (-j)^m \frac{d^m}{dw^m} \bigg|_{w=0} \phi_X(w) $$ i.e. you have to evaluate the $m^{th}$ derivative at $w=0$ to get the $m^{th}$ moment. If you are careful with your summation terms, you will see that in general the $m^{th}$ derivative will look like $$ \sum_{k = p}^{+\infty} (-1)^k \frac{(2k)!}{(2k-2p)!}b^{2k}w^{2k-2p}, \quad m=2p $$ or $$ \sum_{k = p+1}^{+\infty} (-1)^k \frac{(2k)!}{(2k-2p-1)!}b^{2k}w^{2k-2p-1}, \quad m=2p+1 $$ So evaluating the formula at $w=0$ makes all the terms disappear in the case $m=2p+1$ (all the terms contains a $w$ to some positive power) $$ E[X^m] = 0, \quad m \; odd $$ and for $m=2p$, all the terms except the first one i.e. $m=2k$ contains terms with $w$ to some positive power. So at $w=0$, only the first term is left and you get $$ E[X^m] = (-1)^{m/2} (-j)^m m! \: b^{m}, \quad m \; even $$