1a) The reasoning in the first part is right, though not all details are. Let $A$ be the set of sequences that include "the" and $B$ the set of sequences that include "aid." Then
$$|A\cup B|=|A|+|B|-|A\cap B|.$$
For $|A|$, think of "the" as a superletter. Then we have $23$ ordinary letters and a superletter. They can be arranged in $24!$ ways.
A similar calculation deals with $|B|$ and $|A\cap B|$.
1b) We count the number of elements in the complement (either "the" or "math" or both), and subtract from the $26!$ permutations of the alphabet.
Let $P$ be the set of permutations that contain "the," and $Q$ the set of permutations that contain "math." By the reasoning of 1a), $|P|=24!$ and $|Q|=23!$. For $|P\cap Q|$, we need to have the sequence "mathe." There are $22!$ of these.
2) This is left to you for now. Your general strategy was basically OK. But note that for example the total number of permutations of our big word is not $11!$. That would be correct if the letters were all different. However, they are not, so we must divide by $4!4!2!$.
Alternately, there are $\binom{11}{4}$ ways to decide where the I's will go, and for each there are $\binom{7}{4}$ ways to decide where the S's will go, and for each there are $\binom{3}{2}$ ways to decide where the P's will go.
Similar considerations apply to the counting of sequences with restrictions.
You can if you wish colour the letters to make them all distinct, count, and then divide by $4!4!2!$ to deal with the fact they are not all distinct.
1) You are close but you forgot to arrange $A$ and $B$ which you need to multiply again by ${3\choose1}$.
2) You could choose the other four letters instead and ${7\choose3}={7\choose4}$ so the result will be the same. You need not to do both because when you choose one set the other set is automatically chosen as well.
Best Answer
There are $4^5=1024$ sequences where $4^3=64$ are palindromic since such a sequence is determined once the first $3$ bases are determined.
Each of the remaining $960$ is the reverse of one of the other $959$ sequences, so there are $$64 + \frac{960}{2} = 544$$
such sequences.