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.
We denote with
run of length $n$ a substring of $n$ consecutive, equal characters from a binary alphabet $\{0,1\}$.
A maximum run of a string is a run with maximum length.
An $n$-length zero-run is a run consisting of $n$ consecutive $0$s, and a one-run is specified accordingly.
We derive a generating function for the number $a_n$ of strings of length $n$ having no odd, maximum zero-run.
To avoid ambiguities we follow OPs comment and allow odd maximum runs of $1$'s together with odd runs of $0$'s as long as they are less in length than the odd maximum run of $1$'s which seems to be a somewhat more challenging problem.
The string $$0111\color{blue}{000}11$$ has an odd maximum zero-run of length $3$ equal to the maximum one-run and is not admissible. The string $$\color{blue}{0}111011$$ has an odd maximum zero-run of length $1$ less than the maximum one-run of length $3$ and is admissible.
We derive the number of admissible words of length $n$ by counting those words which are not admissible. The number of wanted words is $2^n$ minus this quantity. We do so by deriving a generating function
\begin{align*}
G^{(=2k+1,\geq 2k+1)}(z)\qquad\qquad\qquad k\geq 0
\end{align*}
The first component of the index gives the maximum run length of $0$'s, the second component gives the maximum run length of $1$'s. So, $[z^n]G^{(=2k+1,\geq 2k+1)}(z)$ gives the number of words of length $n$ having maximum zero-runs of length $2k+1$ and maximum one-runs of length less or equal to $2k+1$.
It follows the number $a_n$ of admissible words of length $n$ is
\begin{align*}
a_n=2^n-[z^n]\sum_{k=0}^{\lfloor (n-1)/2\rfloor}G^{(=2k+1,\geq 2k+1)}(z)\qquad\qquad n\geq 0
\end{align*}
We build the generating functions $G(z)$ from the generating function of Smirnov words also called Carlitz words. These are words
with no consecutive equal characters at all. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)
A generating function for the number of Smirnov words over a binary alphabet is given by
\begin{align*}
\left(1-\frac{2z}{1+z}\right)^{-1}
\end{align*}
Replacing occurrences of $0$ in a Smirnov word by one up to $2k+1$ zeros generates words having runs of $0$ with length less or equal $2k+1$.
\begin{align*}
z\longrightarrow z+z^2+\cdots+z^{2k+1}=\frac{z\left(1-z^{2k+1}\right)}{1-z}
\end{align*}
The same can be done with ones and the resulting generating function is
\begin{align*}
G^{(\geq 2k+1,\geq 2k+1)}(z)=\left(1-2\cdot\frac{\frac{z\left(1-z^{2k+1}\right)}{1-z}}{1+\frac{z\left(1-z^{2k+1}\right)}{1-z}}\right)^{-1}
&=\frac{1-z^{2k+1}}{1-2z+z^{2k+1}}
\end{align*}
Since we need a generating function counting words having exactly $2k+1$ zero-runs we calculate
\begin{align*}
&G^{(=2k+1,\geq 2k+1)}(z)=G^{(\geq 2k+1,\geq 2k+1)}(z)-G^{(\geq 2k,\geq 2k+1)}(z)\\
&\qquad=\left(1-\frac{\frac{2z\left(1-z^{2k+1}\right)}{1-z}}{1+\frac{z\left(1-z^{2k+1}\right)}{1-z}}\right)^{-1}
-\left(1-\frac{\frac{z\left(1-z^{2k}\right)}{1-z}}{1+\frac{z\left(1-z^{2k}\right)}{1-z}}
-\frac{\frac{z\left(1-z^{2k+1}\right)}{1-z}}{1+\frac{z\left(1-z^{2k+1}\right)}{1-z}}\right)^{-1}\\
&\qquad=\frac{1-z^{2k+2}}{1-2z+z^{2k+2}}-\frac{(1-z^{2k+1})(1-z^{2k+2})}{1-2z+z^{2k+2}+z^{2k+3}-z^{4k+3}}\\
&\qquad=\frac{(1-z^{2})(1-z^{2k+2})z^{2k+1}}{(1-2z+z^{2k+2})(1-2z+z^{2k+2}+z^{2k+3}-z^{4k+3})}
\end{align*}
We finally conclude
The number $a_n$ of admissible words of length $n$ is
\begin{align*}
a_n&=2^n-[z^n]\sum_{k=0}^{\lfloor (n-1)/2\rfloor}G^{(=2k+1,\geq 2k+1)}(z)\\
&=2^n-[z^n]\sum_{k=0}^{\lfloor (n-1)/2\rfloor}
\frac{(1-z^{2})(1-z^{2k+2})z^{2k+1}}{(1-2z+z^{2k+2})(1-2z+z^{2k+2}+z^{2k+3}-z^{4k+3})}\tag{1}
\end{align*}
The sequence $(a_n)$ starts with
\begin{align*}
(a_n)_{n\geq 1}=(1,2,5,12,24,\color{blue}{48},95,187,367,724,\ldots)
\end{align*}
Example: Let's calculate the number $a_6$ of admissible words of length $6$.
According to (1) we obtain the following series expansions (with some help of Wolfram Alpha)
\begin{align*}
G^{(=1,\leq 1)}(z)&=\frac{(1-z)^2(1-z^2)z}{(1-2z+z^2)(1-2z+z^2)}=\frac{(1+z)z}{1-z}\\
&=z+2z^2+2z^3+2z^4+2z^5+\color{blue}{2}z^6+2z^7+2z^8+2z^9+2z^{10}\cdots\\
G^{(=3,\leq 3)}(z)&=z^3+2z^4+5z^5+\color{blue}{12}z^6+25z^7+53z^8+109z^9+220z^{10}\cdots\\
G^{(=5,\leq 5)}(z)&=z^5+\color{blue}{2}z^6+5z^7+12z^8+28z^9+64z^{10}\cdots\\
\end{align*}
We conclude the number $a_6$ is given by
\begin{align*}
a_6&=2^6-[z^6]\sum_{k=0}^2G^{(=k,\leq k)}(z)\\
&=64-[z^6]\left(G^{(=1,\leq 1)}(z)+G^{(=3,\leq 3)}(z)+G^{(=5,\leq 5)}(z)\right)\\
&=64-\color{blue}{2}-\color{blue}{12}-\color{blue}{2}\\
&=48
\end{align*}
The $64-a_6=16$ non-admissible words are
\begin{array}{cccc}
1\color{blue}{0}1010&\color{blue}{0}10101\\
001\color{blue}{000}&101\color{blue}{000}&011\color{blue}{000}&111\color{blue}{000}\\
\color{blue}{000}100&1\color{blue}{000}10&\color{blue}{000}110&01\color{blue}{000}1\\
11\color{blue}{000}1&\color{blue}{000}101&1\color{blue}{000}11&\color{blue}{000}111\\
1\color{blue}{00000}&\color{blue}{00000}1\\
\end{array}
Best Answer
I confirm your results for $n \le 16$. It might be useful to compute the values by conditioning on $k\in\{1,\dots,\lfloor(n+3)/2\rfloor\}$: \begin{matrix} n\backslash k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 0 & 1 \\ 1 & 1 & 1 \\ 2 & 1 & 3 \\ 3 & 1 & 6 & 0 \\ 4 & 1 & 10 & 2 \\ 5 & 1 & 15 & 8 & 0 \\ 6 & 1 & 21 & 22 & 1 \\ 7 & 1 & 28 & 48 & 6 & 0 \\ 8 & 1 & 36 & 92 & 25 & 0 \\ 9 & 1 & 45 & 160 & 77 & 2 & 0 \\ 10 & 1 & 55 & 260 & 196 & 16 & 0 \\ 11 & 1 & 66 & 400 & 437 & 74 & 1 & 0 \\ 12 & 1 & 78 & 590 & 883 & 254 & 9 & 0 \\ 13 & 1 & 91 & 840 & 1652 & 726 & 54 & 0 & 0 \\ 14 & 1 & 105 & 1162 & 2908 & 1818 & 239 & 2 & 0 \\ 15 & 1 & 120 & 1568 & 4869 & 4116 & 857 & 24 & 0 & 0 \\ 16 & 1 & 136 & 2072 & 7819 & 8602 & 2627 & 156 & 1 & 0 \\ \end{matrix}
Maybe try inclusion-exclusion together with stars-and-bars? For fixed $k$, the first term of inclusion-exclusion is the number of nonnegative integer solutions to $$\sum_{j=1}^k x_j + \sum_{j=1}^{k-1} y_j = n - (k-2) - (k-1) = n-2k+3,$$ which is $$\binom{(n-2k+3) + (2k-1)-1}{(2k-1)-1} = \binom{n+1}{2k-2}.$$ For $k\in\{1,2\}$, this formula is correct. For $k \ge 3$, it is only an upper bound.
An alternative approach is to condition on the tail $(y_{k-1},x_k)$. Explicitly, let state space $$S_n = \left\{k \in \{1,\dots,\lfloor(n+3)/2\rfloor\}, y \in \{[k\not=1],\dots,n\}, x \in \{0,\dots,n-y-2k+5\}\right\}.$$ For $(k,y,x) \in S_n$, let $f_n(k,y,x)$ be the number of such binary strings that end in $1^y 0^x$. Then $f$ satisfies the recursion $$f_n(k,y,x) = \begin{cases} 1 &\text{if $n = 0$} \\ [y = 0 \land x = n] &\text{if $k = 1$} \\ \sum\limits_{\substack(k-1,y_{k-2},x_{k-1}) \in S_{n-y-x}:\\ y_{k-2} \not= y \land ((y_{k-2} \ge 1 \land x_{k-1} \ge 1) \lor k=2)} f_{n-y-x}(k-1,y_{k-2},x_{k-1}) &\text{otherwise} \end{cases}$$
The desired values are then $\sum\limits_{(k,y,x) \in S_n} f_n(k,y,x)$.