Here's a heavy-handed approach. After zero or more rolls, you are in one of four situations:
$\begin{align}
\emptyset:&\qquad\textrm{No 5 or even rolled yet.}\\
E:&\qquad\textrm{Even was rolled, but no 5 yet.}\\
5:&\qquad\textrm{A 5 was rolled, but no even yet.}\\
*:&\qquad\textrm{Both 5 and even have been rolled. Game over.}\\
\end{align}$
The transition matrix of probabilities between each pair of situations is easy to compute:
$\begin{array}{l|cccc}
\nearrow&\emptyset&E&5&*\\
\hline
\emptyset&\frac{1}{3}&\frac{1}{2}&\frac{1}{6}&0\\
E&0&\frac{5}{6}&0&\frac{1}{6}\\
5&0&0&\frac{1}{2}&\frac{1}{2}\\
*&0&0&0&1\\
\end{array}$
So this is now modeled as a absorbing Markov chain with transition matrix
$\left({\begin{array}{cccc}
\frac{1}{3}&\frac{1}{2}&\frac{1}{6}&0\\
0&\frac{5}{6}&0&\frac{1}{6}\\
0&0&\frac{1}{2}&\frac{1}{2}\\
0&0&0&1\\
\end{array}}\right)$
The final state being listed last, the behavior is characterized by the $3\times3$ matrix in the upper left, which is the transition matrix for the non-final states.
$Q=\left({\begin{array}{ccc}
\frac{1}{3}&\frac{1}{2}&\frac{1}{6}\\
0&\frac{5}{6}&0\\
0&0&\frac{1}{2}\\
\end{array}}\right)$
The so-called fundamental matrix $N$ for this chain is
$N=(I-Q)^{-1}
=\left({\begin{array}{ccc}
\frac{3}{2}&\frac{9}{2}&\frac{1}{2}\\
0&6&0\\
0&0&2\\
\end{array}}\right)
$.
The expected number of steps from the $i$-th state to the final one is the sum of the entries of the $i$-th row of $N$, or equivalently the $i$-th entry of the matrix
${\bf t}=N\mathbb{1}=\left({\begin{array}{c}
\frac{13}{2}\\
6\\
2\\
\end{array}}\right)$,
so for the starting state $\emptyset$, it's $\frac{13}{2}$ steps.
The variance of the number of steps from the $i$-th state is the $i$-th entry in the matrix
$(2N-I){\bf t-t_{\textrm sq}}$,
where $t_{\textrm sq}$ is the matrix $\bf t$ with each entry squared. If I didn't slip up with Mathematica,
$(2N-I){\bf t-t_{\textrm sq}}=\left({\begin{array}{c}
\frac{107}{4}\\
30\\
2\\
\end{array}}\right)$,
and the variance you want is $\frac{107}{4}$
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 $\frac{25}{36}\times\frac{1}{36}$
Best Answer
The model is a Markov chain with four states, $0,1,2,3$. The state $0$ is the initial state. The final state is $3$. (The state $k$ means roughly "$k$ matched positions".)
We associate the following constellations to the three states:
(Set delimiters are not really precise, because the roll $1,2,1$ may lead to interpretations when written as $\{1,2,1\}$, but something like $4,*,*$ between set delimiters means a four in some place, then further two no longer useful values (including a possible further occurence of the four, which became useless).)
The passage from one state to the other is given by the scheme:
Let $N_k$ the expected number of rolls needed to pass from the state $k$ to the final state $3$. Then $N_3=0$. Else, $N_k>0$, so we make one step, and get a new state $n$ with the specified passing probabilities, from there we expect $N_n$ steps.
The system is: $$ \left\{ \begin{aligned} N_0 &= 1 + \frac 18N_0+\frac {37}{72}N_1+\frac 13N_2+\frac 1{36}N_3\ ,\\ N_1 &= 1 + \frac 49N_1+\frac 12N_2+\frac 1{18}N_3\ ,\\ N_2 &= 1 + \frac 56N_2+\frac 16N_3\ ,\\ N_3 &=0\ . \end{aligned} \right. $$ We solve this system, the solution is:
$N_3=0$, clear,
$N_2=6$, of course,
$N_1=36/5=7.2$,
$\color{red}{N_0=268/35\approx 7.6571428571428\dots}$ , which is the number asked in the OP.
Some words on the computed passage probabilities $p_{kn}$ from the state $k$ to the state $n$.
Let us simulate. This is the best test. (Sage code is following.)
And this time i've got:
which is not far away from the obtained value $268/35\approx 7.65714285714286\dots$
OK, two more time the same simulation, since i do not like the deviation:
(One may compute the variance using similar methods.)