They are not correct: you do not include the probability of getting to the second roll without having made progress in the overall probability of getting the exact needed roll on the second roll alone.
Which is to say: The probability of making no progress on the first roll and then getting 34 or 43 on the second is $\frac{4}{9} \times \frac{1}{18}$, because you only have a 4-in-9 chance of making no progress.
Similarly, since there's a $\frac{25}{36}$ chance of making no progress on the first roll when the goal is to get two sixes, the probability of getting double six on the second roll after making no progress on the first is $\frac{25}{36}\times\frac{1}{36}$
Representation via generating functions
This isn't satisfactory in the sense that we still cannot obtain a closed form, but the representation is concise and easily programmable. Suppose we have $(k_6, k_8, k_{10}, k_{12})$ dice of types d6, d8, d10, and d12 respectively. Let
\begin{align*}
f_6(x) &= \left(\frac{5}{6}+\frac{1}{6}x\right)^{k_6} \\
f_{8}(x) &= \left(\frac{5}{8}+\frac{2}{8}x+\frac{1}{8}x^2\right)^{k_8} \\
f_{10}(x) &= \left(\frac{5}{10}+\frac{2}{10}x+\frac{2}{10}x^2+\frac{1}{10}x^3\right)^{k_{10}}\\
f_{12}(x) &= \left(\frac{5}{12}+\frac{2}{12}x+\frac{2}{12}x^2+\frac{2}{12}x^3+\frac{1}{12}x^4\right)^{k_{12}} \\
f(x) &= f_6(x)f_8(x)f_{10}(x)f_{12}(x)
\end{align*}
Let $N$ be the random variable denoting the total number of successes (slightly different notation from your post, where you let $N$ represent the value of interest). Then, the probability of getting exactly $n$ successes is
\begin{align*}
P(N = n) =[x^n]f(x)
\end{align*}
where $[x^n]f(x)$ is the coefficient of $x^n$ of $f(x)$. The cumulative distribution function (i.e. the probability of getting $n$ successes or fewer) is
\begin{align*}
P(N \le n) = [x^n]\frac{f(x)}{1-x}
\end{align*}
And so
\begin{align*}
P(N \ge n) = 1 - [x^{n-1}]\frac{f(x)}{1-x}
\end{align*}
Finite-Sample Upper Bound
Let
\begin{align*}
K = k_6 + k_{8} + k_{10} + k_{12}
\end{align*}
and so the proportion of the $K$ dice which are d6, d8, d10, and d12 are respectively
\begin{align*}
(p_6, p_8, p_{10}, p_{12}) = (k_6, k_8, k_{10}, k_{12})/K
\end{align*}
Let $N_k \in \{0, \cdots, 4\}$ ($k = 1, \cdots, K$) be the random variable denoting the success number for each die, and
\begin{align*}
X_m = \sum_{k=1}^{K}\mathbb{I}(N_k = m)
\end{align*}
denote the number of successes produced from the $K$ dice. Then the proportion of the $K$ dice falling in each $m$ ($m = 0, \cdots, 4$), is
\begin{align*}
q_0 &= \frac{5}{6}p_6 + \frac{5}{8}p_8 + \frac{5}{10}p_{10} + \frac{5}{12}p_{12} \\
q_1 &= \frac{1}{6}p_6 + \frac{2}{8}p_8 + \frac{2}{10}p_{10} + \frac{2}{12}p_{12} \\
q_2 &= \frac{0}{6}p_6 + \frac{1}{8}p_8 + \frac{2}{10}p_{10} + \frac{2}{12}p_{12} \\
q_3 &= \frac{0}{6}p_6 + \frac{0}{8}p_8 + \frac{1}{10}p_{10} + \frac{2}{12}p_{12} \\
q_4 &= \frac{0}{6}p_6 + \frac{0}{8}p_8 + \frac{0}{10}p_{10} + \frac{1}{12}p_{12}
\end{align*}
So, $(X_0, \cdots, X_4) \sim \text{Multinomial}(K, (q_0, \cdots, q_4))$.
Finally,
\begin{align*}
P(N \ge n) &= P\left(\sum_{m=0}^{4} mX_m \ge n\right) \\
&= P\left(\exp\left(t\sum_{m=0}^{4} mX_m\right) \ge \exp(tn)\right) & z \mapsto e^{tz} \text{ is increasing for } t>0\\
&\le \frac{E\left[\exp\left(t\sum_{m=0}^{4} mX_m\right)\right]}{e^{tn}} & \text{Markov's inequality} \\
&= e^{-nt}\left(\sum_{m=0}^{4}q_m e^{mt}\right)^K \\
&= \left(\sum_{m=0}^{4}q_m e^{t(m - K^{-1}n)}\right)^K
\end{align*}
and so we can form the Chernoff bounds
\begin{align*}
P(N \ge n) \le \left(\inf_{t>0}\sum_{m=0}^{4}q_m e^{t(m - K^{-1}n)}\right)^K
\end{align*}
Example
Let's suppose we have $(k_6, k_8, k_{10}, k_{12}) = (5, 7, 11, 13)$ and want to find $P(N \ge 30)$. Then
\begin{align*}
P(N \ge 30) = 1 - [x^{29}]\frac{f(x)}{1-x} = 1- \frac{56649270689104302470179125877}{148888471031133469396697088000} \approx 0.6195
\end{align*}
Using the Chernoff bound with
\begin{align*}
K = 36, \mathbf{q} = (0.5405, 0.1931, 0.1456, 0.0907, 0.0301)
\end{align*}
We find that the infimum is attained at $t^* = 0.0894$ giving us $P(N \ge 30) \le 0.8453$.
Best Answer
There are $11$ possible outcomes : $2,3,4,5,6$; $(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$.
The first $5$ of these each have probability $\tfrac16$, and the last six each have probability $\tfrac{1}{36}$.
For each outcome, you have a score and a probability.
The scores are $2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6$.
Probabilities are $\tfrac16$, $\tfrac16$, $\tfrac16$, $\tfrac16$, $\tfrac16$, $\tfrac{1}{36}$, $\tfrac{1}{36}$, $\tfrac{1}{36}$, $\tfrac{1}{36}$, $\tfrac{1}{36}$, $\tfrac{1}{36}$.
To get the average, you multiply the scores by their associated probabilities, and add up. So $$ \frac16(2+3+4+5+6) + \frac{1}{36}(1+2+3+4+5+6) = \frac{141}{36} \approx 3.916 $$