As pointed out in the comments, if you can solve one problem you can solve the other, so I'm just going to give a bound for the first sum. Also, $x > 0$. (It seems like the case $x<0$ might be trickier.) We have
$\begin{align} &\frac{\binom{n}{m}x^m + \binom{n}{m-1}x^{m-1} + \cdots + \binom{n}{0}x^0}{\binom{n}{m}x^m} \\ &= 1 + \frac{m}{(n-m+1)x} + \frac{m(m-1)}{(n-m+1)(n-m+2)x^2} + \cdots + \frac{m!}{n^{\underline{m}}!x^m} \\ &\leq 1 + \frac{m}{(n-m+1)x} + \left(\frac{m}{(n-m+1)x}\right)^2 + \cdots + \left(\frac{m}{(n-m+1)x}\right)^m, \end{align}$
which is the partial sum of a geometric series. Therefore,
$\sum_{k=0}^m \binom{n}{k} x^k \leq \binom{n}{m} x^m \frac{1 - r^{m+1}}{1-r},$ where $r = \frac{m}{(n-m+1)x}.$
Of course, if $r < 1$, then we obtain the simpler expression $\sum_{k=0}^m \binom{n}{k} x^k \leq \binom{n}{m} x^{m+1} \frac{n-m+1}{(n-m+1)x-m}.$
(To give credit where credit is due, this is an adaption of Michael Lugo's answer to a related question on Math Overflow.)