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 $\frac{107}{4}$
