P(HEADS|5) = [P(5 AND Heads)] / P(5)
P(5 and heads) = (1/2)(1/6) = 1/12
P(5) = 1/2 (1/6 + 1/8) = 7/48
(1 / 12) / (7 / 48) = 4 / 7
since you say that we don't know how a head influences the dice choice, I think it is
P(HEADS|5) = [P(5 AND Heads)] / P(5)
P(5) = 1/2 (1/6 + 1/8) = 7/48
P(HEADS AND 5) = (1/2) P(5) = 7 / 96
P(HEADS|5) = (7/96) / (7/48) = 1/2
note P(5) = P(use 8 sided) x (1/8) + P(use 6 sided) x (1/6)
= (1/2) (1/8) + (1/2)(1/6) = 7 / 48
An approach to this problem, a bit lengthy but having the advantage to provide a clear picture,
might be the following.
Start from considering the dice marked.
The set of possible, equi-probable, outcomes is represented by $6^3=216$ triples.
Then consider that you want the outcome of die 2 to be consecutive to the outcome of die 1, while the outcome of die 3 can be whatever
$$
\begin{array}{c|ccc}
{die} & & 1 & 2 & 3 \\
{result} & & {1 \le k \le 5} & {k + 1} & \forall \\
{prob} & & {5/6} & {1/6} & 1 \\
\end{array}
$$
the probability of getting such a scheme is $5/36$.
Now, since in our problem order does not matter, we shall swap (permute) the above configuration.
But we cannot perform that without considering the value of die 3 (call it $j$) compared with the result of die 1 and 2.
In fact, while e.g. $(1,2,3)$ can be permuted in $6$ ways, $(1,1,3)$ can be permuted in $3$ ways.
Moreover, we shall exclude the permuted triples that fall within the range of those already considered.
So, the prospect of the possible ordered configurations and number of ways to swap them is the following
$$ \bbox[lightyellow] {
\begin{array}{*{20}c}
{config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\
\hline
{\left\{ {k,\;k + 1,\;k + 1 < j} \right\}} & {4 + 3 + 2 + 1} & {3!} & {60} \\
{\left\{ {k,\;k + 1,\;k + 1 = j} \right\}} & 5 & 3 & {15} \\
{\left\{ {k,\;k + 1,\;k = j} \right\}} & 5 & 3 & {15} \\
{\left\{ {k,\;k + 1,\;j = k - 1} \right\}} & 4 & 0 & 0 \\
{\left\{ {k,\;k + 1,\;j < k - 1} \right\}} & {1 + 2 + 3} & {3!} & {36} \\
\hline
{{\rm at}\,{\rm least}\,{\rm 2}\,{\rm consecutive}} & {} & {\rm = } & {126} \\
\end{array}
} $$
We see that the fourth configuration is cancelled as being already included in the first.
The prospect for the complementary case (no consecutive outcomes) will give
$$ \bbox[lightyellow] {
\begin{array}{*{20}c}
{config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\
\hline
{\left\{ {k,\;k,\;k} \right\}} & 6 & 1 & 6 \\
{\left\{ {k,\;k,\; \ge k + 2} \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\
{\left\{ {k,\; \ge k + 2,\; = } \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\
{\left\{ {k,\;k + 2,\; \ge k + 4} \right\}} & {2 + 1} & {3!} & {18} \\
{\left\{ {k,\;k + 3,\; \ge k + 5} \right\}} & 1 & {3!} & 6 \\
\hline
{{\rm none}\,{\rm consecutive}} & {} & {\rm = } & {90} \\
\end{array}
} $$
In the case of asking for three consecutive outcomes instead, considering them to be ordered we will have
$$
\begin{array}{c|ccc}
{die} & & 1 & 2 & 3 \\
{result} & & {1 \le k \le 4} & {k + 1} & {k + 2} \\
{prob} & & {4/6} & {1/6} & {1/6} \\
\end{array}
$$
and since each possible triple has distinct values, we can permute them to obtain:
$$ \bbox[lightyellow] { p(\text{3 consec.})={4/6} \cdot {1/6} \cdot {1/6} \cdot 6=1/9=24/216 }$$.
You can verify by direct counting that the values above are correct.
Best Answer
Let $X$ be the sum of the two numbers on the dice. Then $$\Pr[X = x] = \begin{cases}\frac{6-|x-7|}{36}, & x \in \{2, 3, \ldots, 12\}, \\ 0 & \text{otherwise}. \end{cases}$$ Let $Y$ be the number of heads flipped out of $12$ coins. Then $$\Pr[Y = y] = \binom{12}{y} (1/2)^y (1 - 1/2)^{12-y} = \frac{1}{2^{12}} \binom{12}{y}, \quad y \in \{0, 1, \ldots, 12\}.$$ Then the desired probability is $$\Pr[X = Y] = \sum_{x=0}^{12} \Pr[X = x]\Pr[Y = x] = \frac{1}{6^2 2^{12}} \sum_{x=2}^{12} (6 - |7-x|)\binom{12}{x} .$$