d) needs a bit elaboration.
(i) If the question is about a run of exactly 4 $0$'s (though possibly a longer run elsewhere), we can count like this:
There are $2^5$ strings starting with $00001$.
There are $2^5$ strings ending with $10000$.
For $0\le k\le 4$ there are $2^4$ strings $\alpha100001\beta$, where $\alpha$ is $k$ digits and $\beta$ is $4-k$ digits long. Note that we need the "sentinel" $1$s to prevent the block from being longer than 4.
That gives $2^5+2^5+5\cdot2^4=144$ strings.
However, $0000100001$, $0000110000$, $1000010000$ are counted twice, thus the answer is: 141.
Note that e.g. $0000100000$ is counted even though it has (also) a run of 5 0's.
(ii) If the question is about a run of at least 4 $0$'s, we can do this:
There are $2^6$ strings beginning with $0000$, i.e. with the block starting at the first digit.
For $2\le k\le 6$, there are $2^5$ strings of the form $\alpha10000\beta$ with $\alpha$ of length $k-1$ and $\beta$ of length $6-k$, i.e. with the block starting at the $k$th digit.
This give $2^6+5\cdot2^5=224$ strings.
However, we counted solutions with two blocks twice, namely $0000100001$, $0000110000$, $1000010000$, $0000100000$, $0000010000$. (Fortunately, there is not much room for two blocks).
Thus the answer is: 219.
Hint: First, note that since your string has odd length, every binary string has either more 1's or more 0's (no ties). Suppose that a string has more 0's than 1's... what happens when you flip all the bits? The new string has more 1's than 0's. So, for your bijection, consider pairing up a string with it's complement (i.e. the string with all the 1's and 0's flipped).
What you see by this bijection is that for every string with more 1's than 0's there is a corresponding string with more 0's than 1's.
Following your example, what you would pair up are the following:
$000 \to 111$
$001 \to 110$
$010 \to 101$
$100 \to 011$
That accounts for all 8 binary strings of length 3. Note that he LHS are the strings with fewer 1's than 0's and the RHS are the strings with more 1's than 0's.
Best Answer
HINT: Suppose that there are $k$ ones. Then there are $k+4$ zeroes, so $n=2k+4$. In particular, $n$ must be even, so if $a_n$ is the number of acceptable strings of length $n$, then $a_n=0$ when $n$ is odd. Now it’s just a matter of counting the ways to place $k$ ones in a string of length $2k+4$ in such a way that no two of them are adjacent, and at least one of the end characters is a zero.
Think of the ones as dividers; your task is to distribute the $k+4$ zeroes in the $k+1$ spaces before, between, and after the ones in such a way that each of the $k-1$ internal spaces gets at least one zero, and at least one of the end spaces gets a zero. This is a slight variation of a standard stars and bars problem.