First of all, combinations with restrictions are called permutations :)
A neat way to do such problems is to draw as many boxes as there are necessary 'team members' and pencil in the number of permutations for each member. In doing so, you could also draw an arrow protruding from each box where you list the constituent permutations.
In this case, a few sets of boxes can be used as the problem can be split into
disjoint cases.
(Since I don't know how to draw arrows on this site, I'll list the constituent permutations inside the boxes :)
The first case: $\fbox{2 (P. or Ba.)}\fbox{1 (Ba.)}\fbox{3 (G., Bl. or Q.)}\fbox{2 (Bl. or Q.)}\fbox{1 (Q.)}$
Of course, the reindeers could be in a different order within the restrictions of the question — but the same amount of permutations would result. For purposes of visualisation, you can arbitrarily assume which member would go where (within the restrictions of the question).
Make sure, of course, to put the restrictions in first as the permutations for each other position in the 'team' will depend on the restrictions — in this case that Prancer and Balthazar must be next to each other.
The amount of permutations for Case 1? $2*1*3*2*1 = 12.$
There are three other disjoint cases — where Prancer or Balthazar are next to each other in 2nd and 3rd position, 3rd and 4th, or 4th and 5th.
The total number of permutations (the sum of the permutations for each disjoint case) $=48$.
Try it out using this method and see how you go :)
All the books can be arranged in $(2+3+2)!=7!$ ways
There are $3$ branches, three units of books: $\{$History$\}$,$\{$Geography$\}$,$\{$Science$\}$- Arranging branches $=3!$ ways.
Arranging the books within the branches:
History: $2!$
Geography: $3!$
Science:$2!$
Total $=3!(2!\times3!\times2!)=144$ ways
Best Answer
HINT:
$$\underbrace{\underbrace{M_1M_2M_3M_4}_{4!}\mid\underbrace{C_1C_2C_3}_{3!}\mid\underbrace{H_1H_2}_{2!}\mid\underbrace{L_1}_{1!}}_{4!}$$