What do high and low order have a meaning in FIR filter?
1- an FIR filter can be described by the following input/output relation: 
$$ y[n] = \sum_{k=0}^{N} b[k]x[n-k]$$

where $N$ is the order of the FIR filter (its length-1) and those coefficients $b[k]$ of length $N+1$ are the FIR filters impulse response, or the filter coefficients in practice.

As one can see high order FIR filter means $N$ to be large and low order means $N$ to be small. Note that most of those coefficients can be set $0$ but the last one at $x[n-N]$ would still make a high order filter (specifically a filter with a pure delay).

The delay associated with FIR filters (of symmetric impulse response types) is directly related to the order $N$ of the filter. 

2-
An FIR low-pass filter is a mathematical algorithm and its cutoff frequency is normalized to be between $0$ and $\pi$ radians mathematically and between $0$ and $1$ for software such as MATLAB. When you want to find the physical cutoff frequency of such an FIR filter , you should consider the physical sampling rate and relate it through $$f_c = \frac{\omega_c F_s}{ 2\pi} $$
where $\omega_c$ is the cutoff frequency (in radians per sample) of the mathematical filter, and $f_c$ is the physical cutoff frequency in Hertz.

3- Impulse response of the FIR filter is directly available in the coefficients $b[k]$ as I've explained. When you are dealing with an IIR filter or an LTI filter which does not posses a LCCDE expression, then practical computation of the impulse response that can be done in a number of ways. The simplest could be to create a signal that defines the unit impulse as delta = [1, zeros(1,L-1)] then the $L$ points of the impulse response can be computed directly by h = filter(b,a,delta) . Note that if impulse response $h[n]$ does not decay fast enough then you should increase the length $L$.