Let $W(c,n)$ denote the number of words of length $c$ from an alphabet of $n$ letters. Then $W(c,n)=n^c$.
Out of these, the number of words of the same size that do not contain one of the letters is $W(c,n-1)=(n-1)^c$. The number of ways of choosing which letter is missing is $\binom{n}{1}$.
The number of words of the same size that do not contain two letters is $W(c,n-2)=(n-2)^c$. The number of ways of choosing which two letters are missing is $\binom{n}{2}$... and so on ...
Now we use inclusion-exclusion principle: (subtract the number of words missing one of the letters, then add the number missing two of the letters, subtract the number missing three of the letters,...)
We get:
$$W(c,n)-\binom{n}{1}W(c,n-1)+\binom{n}{2}W(c,n-2)-\binom{n}{3}W(c,n-3)+\cdots+(-1)^{n-1}\binom{n}{n-1}W(c,n-(n-1)).$$
This is
$$n^c-\binom{n}{1}(n-1)^c+\binom{n}{2}(n-2)^c-\binom{n}{3}(n-3)^c+\cdots+(-1)^{n-1}\binom{n}{n-1}1^c.$$
or
$$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}(n-k)^c.$$
Another way could be: Denote $S_c^n$ the number of ways to partition the word of length $c$ into $n$ pieces. Then we just need to choose which letter goes to each of the $n$ pieces. This number is $n!$. So the number of words we are looking for is
$$n!S_c^n.$$
The numbers $S_c^n$ are called Stirling's numbers of the second kind.
Your answer for the first part is correct.
How many strings of six lower case letters from the English alphabet contain the letters $a$ and $b$?
You subtracted the number of strings that do not contain $a$ and the number of strings that do not contain $b$ from the total number of strings to obtain $26^6 - 2 \cdot 25^6$. However, you have subtracted the strings that contain neither $a$ nor $b$ twice. Thus, we have to add those $24^6$ strings to the total. By the Inclusion-Exclusion Principle, the number of strings that contain both $a$ and $b$ is $$26^6 - 2 \cdot 25^6 + 24^6$$
How many strings of six distinct lower case letters from the English alphabet contain the letters $a$ and $b$ in consecutive positions with $a$ preceding $b$.
Think of $ab$ as a block. Since the letters are distinct, we must select four other letters from the remaining $24$ letters, which we can choose in $\binom{24}{4}$ ways. We now have five objects to arrange, the block $ab$ and the four other letters we just selected. They can be permuted in $5!$ ways. Hence, the number of strings of six distinct lower case letters from the English alphabet that contain the letters $a$ and $b$ in consecutive positions with $a$ preceding $b$ is $$\binom{24}{4} \cdot 5!$$
How many strings of six distinct lower case letters from the English alphabet contain the letters $a$ and $b$ with $a$ preceding $b$.
Since the letters are distinct, we must choose $4$ of the remaining $24$ letters, which we can do in $\binom{24}{4}$ ways. We can arrange those six letters in $6!$. However, in half of those permutations, $b$ precedes $a$. Thus, the number of six distinct lower case letters from the English alphabet contain the letters $a$ and $b$ with $a$ preceding $b$ is
$$\frac{1}{2} \cdot \binom{24}{4} \cdot 6!$$
Best Answer
HINT
There are $4^n$ different strings of length $n$ using $a$, $b$, $c$, or $d$.
So, if there are $f(n)$ strings of length $n$ with have an even number of $a$'s, then that means that there are $4^n-f(n)$ strings of length $n$ with an odd number of $a$'s.
I don't want to give it completely away, but this is a big hint ... I think with this you'll be able to complete the inductive step no problem ...