Each part has to be odd, so the series whose exponents describe the possibilities for one part is
$$x+x^3+x^5+\ldots=x\sum_{n\ge 0}x^{2n}\;.\tag{1}$$
You want $k$ parts, and so you want
$$\left(x\sum_{n\ge 0}x^{2n}\right)^k\;:$$
each term of that $k$-th power is of the form $$x^{2n_1+1}x^{2n_2+1}\ldots x^{2n_k+1}\;,\tag{2}$$ where the $i$-th factor corresponds to the $i$-th part of the composition. In other words, the term $(2)$ in the power corresponds to the composition
$$(2n_1+1)+(2n_2+1)+\ldots+(2n_k+1)\tag{3}$$
of the number $n$ that is the actual sum in $(3)$.
At this point you should be able to find the closed form generating function for $(1)$ and then take its $k$-th power. When it comes to extracting the coefficient of $x^n$, you’ll probably find it useful to know the series expansion of $\frac1{(1-x)^k}$; if you don’t already know it, it can be obtained by repeated differentiation of one of the most familiar ones.
Ad c.)
Your approach is fine. With ${\mathcal{K}} = \{1,2\}^{\ast}$ we obtain
\begin{align*}
K(z) &= \frac{1}{1-\left(z+z^2\right)}\\
&=\frac{1}{1-z-z^2}\tag{1}
\end{align*}
and with $\tilde{\mathcal{K}} = \{2,3,4,5,6,...\}^{\ast}$ we obtain
\begin{align*}
\tilde{K}(z) &= \frac{1}{1-(z^2+z^3+z^4+z^5+\cdots)}\\
&=\frac{1}{1-\frac{z^2}{1-z}}\tag{2}\\
&=\frac{1-z}{1-z-z^2}\\
&=1+\frac{z^2}{1-z-z^2}\tag{3}
\end{align*}
Comment:
We obtain from (1) and (3) for $n\geq 1$
\begin{align*}
\color{blue}{[z^{n+2}]\tilde{K}(z)} &=[z^{n+2}]\left(1+\frac{z^2}{1-z-z^2}\right)\\
&=[z^{n+2}]\frac{z^2}{1-z-z^2}\tag{4}\\
&=[z^n]\frac{1}{1-z-z^2}\tag{5}\\
&\,\,\color{blue}{=[z^n]K(z)}
\end{align*}
and the claim follows.
Comment:
Note the coefficients of
\begin{align*}
K(z)&=\frac{1}{1-z-z^2}\\
&=\color{blue}{1}+\color{blue}{1}z+\color{blue}{2}z^2+\color{blue}{3}z^3+\color{blue}{5}z^4+\color{blue}{8}z^5+\cdots
\end{align*}
are the Fibonacci numbers.
ad a.)
We start with binary sequences with no consecutive equal characters at all. See example III.24 Smirnov words from Analytic Combinatorics by P. Flajolet and R. Sedgewick for more information.
A generating function for the number of Smirnov words over a binary alphabet is given by
\begin{align*}
\left.\left(1-\frac{u}{1+u}-\frac{w}{1+w}\right)^{-1}\right|_{u=w=z}\tag{6}
\end{align*}
where $u$ represents occurrences of $0$ and $w$ occurrences of $1$.
Since there are no restrictions stated for zeros we replace occurrences of $0$ in a Smirnov word by one or more zeros.
\begin{align*}\
u\longrightarrow u+u^2+u^3+\cdots=\frac{u}{1-u}\tag{7}
\end{align*}
We substitute (7) in (6) evaluate at $z$ and obtain
\begin{align*}
\left(1-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}-\frac{z}{1+z}\right)^{-1}
&=\left(1-z-\frac{z}{1+z}\right)^{-1}\\
&=\frac{1+z}{1-z-z^2}\\
&=1+2z+3z^2+5z^3+8z^4+\cdots
\end{align*}
which is again a generating function of (shifted) Fibonacci numbers.
Best Answer
Let $\mathcal{O}$ denote the class of all odd numbers, so that it has generating function $O(z) = z + z^3 + z^5 + \dots = \dfrac{z}{1-z^2}$. Then the class of compositions into odd parts is $$\mathcal{G} = \operatorname{S\scriptsize EQ}(\mathcal{O}) \implies G(z) = \frac{1}{1-O(z)} = \frac{1}{1-\frac{z}{1-z^2}} = \frac{1-z^2}{1-z-z^2}$$ where $G(z) = \sum_{n \ge 0} g(n) z^n$ is the generating function for $\mathcal{G}$.
Similarly, let $\mathcal{C}$ denote the class containing just the numbers $1$ and $2$ (so $C(z) = z+z^2$), then the class of compositions into parts equal to $1$ and $2$ is $$\mathcal{H} = \operatorname{S\scriptsize EQ}(\mathcal{C}) \implies H(z) = \frac{1}{1-C(z)} = \frac{1}{1-z-z^2}$$ where $H(z) = \sum_{n \ge 0} h(n) z^n$ is the generating function for $\mathcal{H}$.
Now you want to prove that $g(n) = h(n-1)$ for $n \ge 1$, or equivalently that $g(n+1) = h(n)$ for $n \ge 0$. We have $$\sum_{n \ge 0}g(n+1)z^n = \frac{G(z) - g(0)}{z} = \frac1z \left( \frac{1-z^2}{1-z-z^2} - 1\right) = \frac{1}{1-z-z^2} = H(z)$$ which proves the assertion.