Numbers of ways to buy a dozen bagels of various types, and eat them Three types of bagels: honey, blueberry, and sesame seed. Assume that
bagels of each type are indistinguishable.
(a) How many ways are there to buy a dozen bagels?
$(3 + 12 - 1) \choose 12$
because choosing 12 distinguishable from 3 distinguishable
(b) How many ways are there to buy a dozen if at least 2 of each type of
bagel are chosen?
$(3 + 6 - 1) \choose 6$
same as part a 
(c) Suppose you bought 4 of each kind, as well as a tub of cream cheese.
How many different orders are there in which to eat the bagels, if in
addition you have the option of putting cream cheese on each bagel?
not sure
 A: These first two can be done by the star and bar method, which I think you have done correctly.
a)$$x_h+x_b+x_s=12,x_i\ge0\\\binom{12+3-1}{3-1}=91$$
b)$$x_h+x_b+x_s=12,x_i\ge2\\\binom{12-2\times3+3-1}{3-1}=28$$
The third can be done by arranging the 12 bought bagels in a row $12!$, the position from one representing the order in which to eat, note that each group of 4 bagels is identical hence we decide by $4!$ for each group.Now we have just two choices each bagel, either it will have cream cheese or not, which leaves 2 ways for each 12 bagels so we multiply by $2^12$.
c)$$\frac{(4\times3)!}{4!^3}\times2^{4\times3}\approx1.41\times10^8\tag{$141926400$ to be precise}$$
A: a) Your expression for the answer is correct.
b) In effect you are buying $6$ bagels. Then use the same method as in a). The answer will be quite a bit  smaller than the answer for a).
c) There are $\frac{12!}{4!4!4!}$ (multinomial coefficient) ways to arrange the flavours. 
Alternately, you can choose the location of the flavour A bagels in $\binom{12}{4}$ ways, and for each way choose the location of the flavour B bagels in $\binom{8}{4}$ ways.
For each arrangement of flavours, there are $2^{12}$ ways to make the cream cheese, no cream cheese decisions. 
