Added 2016-03-13: This answer is just a starter to the second part of OPs question, namely finding words over the ternary alphabet $\{0,1,2\}$ having maximum run length $k$ and containing the string $1^k2$ but no other strings with $1$ and run length $k$.
A detailed answer based upon a two-step language decomposition
\begin{align*}
\left(\varepsilon+\mathcal{H}_0+\mathcal{H}_2\right)&\left(11^{<k-1}\left(\mathcal{H}_0+\mathcal{H}_2\right)
+1^k\mathcal{H}_2\right)^*1^{<k}\\
&\qquad\qquad\text{with}\\
&\mathcal{H}_0=00^{<k}\left(22^{<k}00^{<k}\right)^*2^{<{k+1}}\\
&\mathcal{H}_2=22^{<k}\left(00^{<k}22^{<k}\right)^*0^{<{k+1}}\\
\end{align*}
and resulting in a generating function
\begin{align*}
A_k(z)&=(1+2H(z))\sum_{q=0}^\infty\left(z\frac{1-z^{k-1}}{1-z}2H(z)+z^kH(z)\right)^q\frac{1-z^k}{1-z}\\
&=\frac{\left(1-z^k\right)\left(1-z^{k+1}\right)}{1-3z+2z^{k+1}+2z^{k+2}-z^{2k+1}-z^{2k+2}}
\end{align*}
is stated in this MSE question.
Note: We derive the number of words built from $\{0,1,2\}$ having a maximal run of $1$ of length $k$ and no other maximal runs of length $\geq k$. We do not consider the additional constraint that the maximal run of $1$ has to be followed by $2$.
This approach is based upon example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick.
Smirnov Words: Words without runs
Smirnov Words are words having no consecutive equal letters. They can be related to unconstrained words and vice versa in the following way:
If an unconstrained word is given, we can collaps each run of consecutive letters into a single letter associating a Smirnov word this way. Conversely, starting from a Smirnov word and substituting each letter by a sequence of length $\geq 1$ of this letter we can get all unconstrained words.
Arbitrary words are derived from Smirnov words by a simultaneous substitution:
\begin{align*}
W(v_1,\ldots,v_c)=S\left(\frac{v_1}{1-v_1},\ldots,\frac{v_c}{1-v_c}\right).\tag{1}
\end{align*}
Relation (1) determines the generating functions $S$ for Smirnov words implicitely. Since the inverse function of $\frac{v}{1-v}$ is $\frac{v}{1+v}$ we find:
\begin{align*}
S(v_1,\ldots,v_c)=W\left(\frac{v_1}{1+v_1},\ldots,\frac{v_c}{1+v_c}\right)
=\left(1-\sum_{j=1}^{c}\frac{v_j}{1+v_j}\right)^{-1}
\end{align*}
Hint: A more detailed explanation of Smirnov words is given in this MSE answer
Words with runs of length $< k$
Let's consider the alphabet $\{v_0,v_1,v_2\}$. Starting with a Smirnov word we can build from it words with runs of length $<k$ by substituting
$$v_j\mapsto v_j+\cdots+v_j^{k-1}=v_j\frac{1-v_j^{k-1}}{1-v_j}$$
Replacing $v_j$ with $z$ we obtain a generating function describing all words built from $\{0,1,2\}$ having a run of length $<k$
\begin{align*}
G^{<(k,k,k)}(z)&=\left(1-3z\frac{1-z^{k-1}}{1-z^k}\right)^{-1}=\frac{1-z^k}{1-3z+2z^k}
\end{align*}
In order to find the number of words with a maximal run of $1$ of length $k$ and no other runs with a length $\geq k$ we use
\begin{align*}
G^{<(k,k+1,k)}(z)&=\left(1-2z\frac{1-z^{k-1}}{1-z^k}-z\frac{1-z^{k}}{1-z^{k+1}}\right)^{-1}\\
&=\frac{1-z^k-z^{k+1}+z^{2k+1}}{1-3z+z^k+z^{k+1}+2z^{k+2}-2z^{2k+1}}
\end{align*}
We conclude: The number of words of length $n$ from the alphabet $\{0,1,2\}$ with a maximum run of $1$ of length $k$ and with maximum runs of $0,2$ of length $<k$ is the coefficient of $z^n$ of
\begin{align*}
G^{<(k,k+1,k)}(z)-G^{<(k,k,k)}(z)
\end{align*}
$$ $$
Example $k=3$
With the help of Wolfram Alpha we obtain
\begin{align*}
G^{<(3,3,3)}(z)&=\frac{1-z^3}{1-3z+2z^3}\\
&=1+3z+9z^2+24z^3+66z^4+180z^5+\mathcal{O}(z^6)\\
G^{<(3,4,3)}(z)&=\frac{1-z^3-z^4+z^7}{1-3z+z^3+z^4+2z^5-2z^7}\\
&=1+3z+9z^2+25z^3+70z^4+196z^5+\mathcal{O}(z^6)\\
\end{align*}
The difference of the generating functions is
\begin{align*}
G^{<(3,4,3)}(z)-G^{<(3,3,3)}(z)=z^3+4z^4+16z^5+\mathcal{O}(z^6)
\end{align*}
and the words with runs of $1$ of length $k=3$ are
\begin{align*}
z^3\quad&\quad111\\
z^4\quad&\quad0111\quad2111\quad1110\quad1112\\
z^5\quad&\quad00111\quad02111\quad10111\quad12111\quad20111\quad22111\\
&\quad01112\quad21110\quad01110\quad21112\\
&\quad11100\quad11101\quad11102\quad11120\quad11121\quad11122
\end{align*}
