To count the number of 8-bit strings that have exactly 3 zeros and at least 2 are adjacent, you can count the number of 8 bit strings with exactly 3 zeros, and subtract those in which they are not adjacent.
The number of 8-bit strings with exactly 3 zeros is $\binom{8}{3}$. To count those with no two zeros adjacent, consider placing the five $1$s, leaving space between them for possibly adding a $0$:
$$\Box \quad1\quad \Box\quad 1 \quad\Box \quad 1 \quad \Box\quad 1 \quad \Box \quad 1\quad \Box$$
Now you will select three of the six boxes to place a zero in. There are $\binom{6}{3}$ ways of doing this. So the number of $8$-bit strings with exactly $3$ zeros, and at least two zeros adjacent is
$$\binom{8}{3} - \binom{6}{3} = 56 - 20 = 36.$$
I don't know how you counted for $4$. Doing the same thing as above, there are $\binom{8}{4}$ strings with exactly four $0$s; if you place all the $1$s first, there are five places where a zero may go, so you need to subtract $\binom{5}{4}$ strings that have no two zeros adjacent; that gives $\binom{8}{4}-5 = 65$ strings.
The rest seem okay.
Using these totals, we have:
$0 + 7 + 36 + 65+ 56+28+8+1=201$ total.
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.
Best Answer
Hint: Use a recurrence relation. What if the string with $k$ bits starts with a 1, how many possibilities gives this for strings with $k+1$ bits? If the string with $k$ bits starts with a 0, how many possibilities gives this for strings with $k+1$ bits?