I'll kick off with some observations.
I think there is no easily determinable winning strategy. The game as envisaged is mind-blowing because there are 4,263,421,511,271 possible ways of arranging your army. However, suppose that there were a deterministic configuration that maximized either a or b. Then both sides would use it and all 10 platoons would be a draw.
But, if you knew what your opposition were going to do, then you'd just mirror their army, take the smallest platoon with size $\geq 9$, and give 1 to each other army winning 9 battles.
Then again, it would depend on what your opponent's objectives were. If he simply wanted to save as many men as possible, he'd do something like $(100,0,0,\ldots)$. Easy to beat with objective (a) and impossible with objective (b). So it is important to know whether the opposing general knows about your objective and/or has an objective of their own.
But normally this reduces to something like rock, paper, scissors where the Nash Equilibrium solution is to have an equal probability of choosing rock, paper, or scissors.
Determining a Nash Equilibrium for such a large solution space is not trivial. So here are some numerical attempts for much simpler problems:
For both problems, I assume that defeat means you need your platoon to be strictly bigger than the opposing platoon. I then assume 9 soldiers and 3 platoons, and evolve the strategy until I'm roughly indifferent to what my opponent is choosing. I also assume that my probabilities are going to be equal for all permutations. So whatever my probability is for [0,3,6] it will be the same for [0,6,3], [3,0,6], and so on.
For problem a, it shows the following (with a lot of spurious accuracy!):
[0 0 9] 0.08969140375863138
[0 2 7] 0.033696054860265535
[0 4 5] 0.029747584762232965
[0 3 6] 0.027726778078327265
[0 1 8] 0.025936954631579415
[1 1 7] 0.006435624011966441
[1 4 4] 0.002057572668762006
[3 3 3] 0.0018396498692940773
[2 2 5] 0.0003207716042842319
[1 2 6] 4.886082454138541e-13
[2 3 4] 4.0188092068566764e-13
[1 3 5] 1.124292366185872e-19
I assume I've made an error, since the high probability of the [0,0,9] solutions seems wrong (It might be correct since it gives a high probability of a score of 1 when the max score in this case is 2). But I'll post this for now to show the approach. The score you would get in this case (against any choice from player 2) is $1.18\pm 0.01$. Of course, your max score is 2 and your min is 1.
Here's the case for 6 soldiers and 4 platoons showing similar trends:
[0 0 0 6] 0.03507904334840038
[0 0 2 4] 0.029695480610278505
[0 0 3 3] 0.02956549030935802
[0 0 1 5] 0.02715661099230036
[1 1 1 3] 1.3911109098506814e-05
[0 2 2 2] 1.362817469650331e-06
[1 1 2 2] 7.786961895140738e-07
[0 1 1 4] 1.469657831039888e-09
[0 1 2 3] 1.1765812916837055e-18
I'm still having troubles convincing myself that I haven't made an error. I mean, you basically always get 1 unless you have the exact same strategy, so why the preference for the [0,0,0,6] type solutions? Here the average score is $1.35\pm0.03$
For problem b, I don't think there is one because you will always get 0 if your opponent plays [0,0,9] as discussed earlier.
Another observation: Doing the 9 soldiers with 3 platoons but setting the [0,0,9] solutions to 0 probability yields:
[0 2 7] 0.051643285123016455
[0 4 5] 0.04878474186500994
[0 3 6] 0.03799226690918427
[1 1 7] 0.022924736330020803
[0 1 8] 0.011810956237999548
[1 4 4] 0.007329407508318203
[3 3 3] 0.006553131156166946
[2 2 5] 0.00043231217051718226
[2 3 4] 9.779932406154017e-13
[1 2 6] 2.2520149653133117e-14
[1 3 5] 4.0049117381710107e-19
[0 0 9] 0.0
It's far further from an equilibrium. But interestingly, the results are generally higher scores. The score ranges from $1.24$ to $1.4$ with an average of $1.28$. Something weird is going on...
I'll be interested to see what other people come up with.