How does variance in task completion time affect makespan?
As $m = k \times n$, we can look at this in terms of $k$ and $n$ instead of $n$ and $m$. Let's say $T_i$ is the time it takes the $i$-th processor to finish its work.

As $n$ grows, the probability that $T_i$ = $5k$ (the processor was assigned only $T=5$ tasks) for some $i$ approaches $1$, so makespan being defined as $\mathrm{max}(T_i)$, $E[M]$ approaches $5k$.

For the second scenario this is $4k$ so increasing the number of processors makes the 4–2 split better.

What about $k$ — increasing the number of tasks per processor? Increasing $k$ has the opposite effect, it makes it less likely to have a processor with an unlucky set of tasks. I'm going home now but I'll come back to this later. My "hunch" is that as $k$ grows, the difference in $E[M]$ between the 4–2 split and the 5­­­–1 split disappears, $E[M]$ becomes the same for both. So I would assume that 4–2 is always better except maybe for some special cases (very small specific values of $k$ and $n$), if even that.

So to summarize:


Lower variance is better, all else being equal.
As the number of processors grows, lower variance becomes more important.
As the number of tasks per processor grows, lower variance becomes less important.