Counting exercises - Solution verification. i'm studying some combinatorics and i came up in the following exercises. Suppose we are given a set $U$ of $n$ elements. Suppose $A \subset U$ has $k$ elements. Determine the number of subsets $B\subset U$ satisfying:

$i)$ $B\subset A$,
$ii)$ $B\supset A$,
$iii)$ $A\cap B = \emptyset$
$iv)$ $A \cap B \neq \emptyset$
Determine further
$v)$ How many pairs of subsets $(C,D)$ of $U$ can you find satisfying the strict inclusion $D \subset C$?

I tried to find a solution to all these problem by myself, but i'm not sure whether it is right or wrong and i would be very thankful to anyone who corrects me my mistakes.
Solutions
$i)$ I've thought that since $A$ has $k$ elements each element in $A$ can belong to an hypothetical set $B\subset A$ or not. Therefore i've thought that the total number of set $B\subset U$ satisfying $B\subset A$ is
$$2^k$$
$ii)$ In this case $B$ must contain all the $k$ elements contained in $A$, therefore the other $n-k$ element contained in $U\backslash A$ can belong to $B$ or not, but in each case we would have that the condition $B\supset A$ is satisfied. So there are $$2^{n-k}$$
number of subset $B$.
$iii)$ This time $B$ can't contain any element contained in $A$, but we still can choose whether each element contained in $U\backslash A$ belongs to $B$ or not, therefore there are $$2^{n-k}$$
possible choices for $B$.
$iv)$ Suppose we only want an element of $A$ to belong in $B$ then we can choose this element in ${k\choose 1}$ different ways. Once we have blocked this element we can choose the other $n-k$ element belonging to $U\backslash A$ to belong in $B$ or not. Therefore if we want only an element in $A$ to belong in $B$ we can find $${k\choose 1}2^{n-k}$$ subsets $B$ such that $A\cap B \neq \emptyset$. Suppose we want now two elements of $A$ to belong in $B$. We can choose $${k\choose 2}2^{n-k}$$ different subsets $B$ such that $A\cap B \neq \emptyset$ and so on. The final answer is therefore that $B$ can be chosen in
$${k\choose 1}2^{n-k} + {k\choose 2}2^{n-k} + \cdots + {k\choose k}2^{n-k}$$
different ways.
$v)$ Since the inclusion is strict i began by counting the number of subsets $C$ containing two elements, that is ${n\choose 2}$ For each subset we can find ${2\choose 1}$ subsets $D$ such that the inclusion $D\subset C$ is strict. Then i've counted the number of subsets $C \subset U$ containing $3$ elements, that is ${n\choose 3}$. For each of these subset we can find ${3\choose 1}+{3\choose 2}$ subsets $D$ such that the inclusion $D\subset C$ is strict. Therefore the final answer is
$${n\choose 2}{2\choose 1} + {n\choose 3}\left({3\choose 1}+{3\choose 2}\right) + \cdots + {n\choose n}\left({n\choose 1}+\cdots + {n\choose n-1}\right)$$

There is another problem that i can't really solve and i would really appreciate any hints on it. The problem is the following

$vi)$ In how many ways can you select two not necessarily distinct subsets of $U$ so that their union is $U$? (the order doesn't count)

 A: The first three answers are right, and the reasoning is well explained.
For $A\cap B\ne \emptyset$, your approach is good, but there is a much simpler way. There are $2^n$ subsets, and (by the previous problem) $2^{n-k}$ that have empty intersection with $A$, so there are $2^n-2^{n-k}$ with non-empty intersection with $A$.
For the fifth problem, we want strict inclusion. Forget temporarily about the strict part. This is not an issue, for there are $2^n$ choices where $C=D$, and they can be removed later. We need to divide $U$ into $3$ parts, the part in $D$, the part in $C\setminus D$, and the rest. So for every element of $U$, we have $3$ choices, for a total of $3^n$. Now subtract $2^n$.
For the last problem, the basic idea is much the same. We want to express our set as a union of sets $A$ and $B$. Essentially, we want to divide $U$ into three parts, $A\setminus B$, $B\setminus A$, and $A\cap B$. If we are dividing into labelled parts, there are $3^n$ ways to do the job. But the choice $A=B=U$ is special. Forget about that one temporarily. Since labels do not matter, the $3^n-1$ possibilities apart from the special one should be divided by $2$. Now add back the special one. 
A: Your solutions for $i)$, $ii)$ and $iii)$ are good. Simple enough, right?
Your solutions for $iv)$ and $v)$ are correct but messy. I wonder if there isn't a cleaner way to write them? What is the sum of entries in a row of Pascal's Triangle?
For $vi)$ you can approach it this way: If the order of $U$ is $n$, you will have one subset $A$ containing $k$ elements, and your other subset $B$ must contain the remaining $n-k$ elements, but it may also contain any of the elements of $A$.
