I am not sure of my result.
First we define $X$ and $X=k+1$ when in $k+1$-throw we obtain 5 or 6 for the first time. Then $P(X=k+1)=(4/6)^k(2/6)$ and $\mathbf{E}X=\sum_{k=0}^{\infty}(k+1)P(X=k+1)=3$ So now we have 5 or 6 and we need to wait for the other number. So we start "from the beginning" but now we wait for one specific number, so expected value is $\mathbf{E}Y=\sum_{k=0}^{\infty}(k+1)P(Y=k+1)=6$ where $P(Y=k+1)=(5/6)^k(1/6)$. So our expected amount of throws is $9$.
Before looking for variance I'd like to confirm that result.