A linear system is inconsistent is if it represents a contradiction, for instance the system
$\left[\begin{array}{cc|c} 0 & 0 & -10 \\ 3 & -2 & 1 \\ \end{array}\right]$
is inconsistent because the first line represents a linear equation $0x+0y=-10$, i.e. $0=-10$, which is a contradiction. Geometrically, when you solve a 2x2 linear system, you are finding the intersection between a pair of lines. If you reach a contradiction, like the system above, then your lines do not intersect, i.e. they must be parallel.
If you are being asked this question, you have probably already covered Gauss-Jordan ellimination. Inconsistencies in linear systems can be readily identified if the system is brought to reduced row echelon form (can you see why?), so I would start with that. The steps are simple:
$\left[\begin{array}{cc|c} 1 & h & -5 \\ 2 & -8 & 6 \\ \end{array}\right]$ Multiply the second row by $1/2$: $\left[\begin{array}{cc|c} 1 & h & -5 \\ 1 & -4 & 3 \\ \end{array}\right]$ Subtract the second row from the first: $\left[\begin{array}{cc|c} 0 & h+4 & -8 \\ 1 & -4 & 3 \\ \end{array}\right]$ Without even proceeding further, it is obvious that one way for the system to be inconsistent is if the first line is $0\, 0\, |\, -8$, since this would be equivalent to saying $0x+0y=-8$, that is $0=-8$, a contradiction. The first row would have this form only if $h=-4$, so $h=-4$ makes the system inconsistent.
Now it is pretty clear at this point that no other value of $h$ would make the system inconsistent, and after you are comfortable with Gauss-Jordan elimination this fact would be apparent to you as well, though you should really try to understand why first. So let's say $h\ne-4$. Then we can multiply the first row by $\frac 1 {h+4}$: $\left[\begin{array}{cc|c} 0 & 1 & -\frac 8 {h+4} \\ 1 & -4 & 3 \\ \end{array}\right]$ And now add 4 times the first row to the second: $\left[\begin{array}{cc|c} 0 & 1 & -\frac 8 {h+4} \\ 1 & 0 & 3-\frac {32} {h+4} \\ \end{array}\right]$ To really be precise, you can swap the two rows: $\left[\begin{array}{cc|c} 1 & 0 & 3-\frac {32} {h+4} \\ 0 & 1 & -\frac 8 {h+4} \\ \end{array}\right]$
Thus for any value of $h$ other than $-4$, we can solve the system - there is no way to make the system displayed above have a row which looks like $0\,0\,|\,c$, for any non-zero number $c$.