Solution verification about equivalence classes. 
I'm absolutely not sure about the notations and the correct terminology and I was wondering if everything checks out..
Question 5
a.
$4^{2\ }-1$ = 15
Therefore we have 15 different partitions.
b.
The number of partition is the same as the number of equivalence relations.
Therefore we have 15 different relations.
c.
By limitation of being an equivalent relation, we have three independent partitions in $\left\{\left\{7,4,2\right\},\left\{1.3\right\},\left\{5\right\}\right\}$
with $\left\{6\right\}$ free to join any partition, or be a partition on it's own.
Therefore we have 4 equivalence relations that satisfy the condition.
The largest one is:
$R=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right),\left(7,7\right),\left(7,4\right),\left(4,7\right),\left(7,2\right),\left(2,7\right),\left(7,6\right),\left(6,7\right),\left(4,2\right),\left(2,4\right),\left(2,6\right),\left(6,2\right),\left(3,1\right),\left(1,3\right),\left(6,4\right),\left(4,6\right)\right\}$
 A: My check on the answers with discussion:
a) How many different partitions of the set $\{3,19,27,40\}$ are there?
As JMoravitz comments, the number of partitions is given by the Bell numbers, but for a set of size $4$ it's not difficult to work out de novo, by largest part size $P_M$:
$P_M = 1$: All parts are size $1$, #options $=1$.
$P_M = 2$: Other parts size $1$: #options $=\binom 42 = 6$.
$\phantom{P_M = 2}$: Both parts size $2$ (symmetric): #options $=\frac12\binom 42 = 3$.
$P_M = 3$: Other part must be size $1$: #options $=\binom 43 = 4$.
$P_M = 4$: Whole set is one part, #options $=1$.
Total partitions: $15$
b) How many different equivalence relations can be defined on this set?
As you stated, each equivalence relation can be mapped to & from a partition of the set, so again $15$.
c) How many equivalence relations on $\{1,2,3,4,5,6,7\}$ contain the set $\{(7,4),(6,6),(1,3),(4,2)\}$ but are disjoint from the set  $\{(5,4),(7,3),(3,5)\}$?
The partition implied by the first condition implies that one part has a subset of $\{2,4,7\}$  and one part has a subset of $\{1,3\}$. The exclusions given in the second condition imply that these two parts are distinct and that $5$ is also not a member of either of them, giving us a third part that has $\{5\}$ as a subset. So the only remaining choice is where to put $6$, which is in any of the three parts identified or in a new part of its own, giving $4$ different equivalence relations.
- Out of the equivalence relations for which the conditions hold, [state which] has the largest number of pairs in it.
Adding $6$ as an isolated equivalence class,  $\{\{2,4,7\},\{1, 3\},\{5\},\{6\}\}$ would give the $7$ reflexive pairs with $6+2$ more pairs in the first two classes. Combining $6$ into another class would add pairs in proportion to the pre-existing size of that class, meaning that combining it into the largest class will give most pairs: $\{\{2,4,6,7\},\{1, 3\},\{5\}\}$ which has $7+12+2=21$ pairs.
