For $a.1$, there are $$\frac {11*10}{2}=55$$ ways to pick the first permutation. If you just prohibit inverses(the inverse is the same as the swap), there are then $$\frac {55*54*53*52}{24}$$ ways to pick four permutations. But this ignores the fact that you may move the same letter twice. If you want eight letters to move in four swaps, you have $$\frac {55*36*21*10}{24}$$
I will give a basic counting solution. If you want an answer using generating functions, look at the answer to this question.
If all five letters are different (which happens when you use one of each), you get $5!=120$ words.
If only two of the letters are equal (two 'S', two 'T', or two 'I', and $\binom{4}{3}$ ways to choose the remaining three), you have $\frac{5!}{2!}=60$ words for each of {'S', 'T', 'I'}. In total $720$ words.
If you have "two pairs" (there are three ways to get the pairs, and three ways to get the last letter in each case), there are $\frac{5!}{2!2!}=30$ words. In total for all these cases, there are $270$ words.
If you have "full house" (there are two ways to get three equal, and two ways for each of those to get the last pair), there are $\frac{5!}{3!2!}=10$ words. In total for these cases, there are $40$ words.
If you have three equal and two different (the triple can be had in two ways, and the remaining two can be had in $\binom{4}{2}=6$ ways for each triple), there are $\frac{5!}{3!}=20$ words. For all these cases, there are $240$ words.
Summing all of these cases gives $1390$ words.
Best Answer
M=1
I=2
S=4
A=2
U=1
G=1
Case 1: All 4 letters are same
There is only one arrangement for this.
Case 2: 3 letters are similar
The three letters must be S. The remaining letter can be chosen in ${5 \choose 1}$ ways and these can be arranged in $\frac {4!}{3!}$ ways. This makes a total of 20 arrangements.
Case 3: 2 pairs are similar
The two repeating letters can be chosen in ${3 \choose 2}$ ways and can be arranged in $\frac {4!}{{2!}{2!}}$. This equals 18 arrangements.
Case 4: 2 are similar
The repeating letters can be chosen in ${3 \choose 1}$ ways and the remaining two in ${5 \choose 2}$ ways. These can be arranged in $\frac {4!}{2!}$ ways. This equals 360 arrangements.
Case 5: All 4 are different The 4 letters can be chosen in ${6 \choose 4}$ ways and can be arranged in $4!$ ways. This equals 360 ways.
Hence the total number of arrangements is $1+20+18+360+360=759$.