$\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\E}{\mathrm{E}}$
Not sure if I have done this correctly because the answer seems rather large to me.
Define $X$ to be $0$ if the first toss is tails, 1 if the first toss is heads and the second is tails, and 2 if the first toss is heads and the second is heads.
I have already solved the $\E[N] = 6$.
Then we have $\Var[N] = \E[N^2] - (E[N])^2.$ Focusing on $\E[N^2]$ we get:
$\E[N^2] = \E[\E[N^2|X]] = \E[N^2|X=0]P(X=0) + \E[N^2|X=1]P(X=1) + \E[N^2|X=2]P(X=2)$.
If $X = 0$, then we can know that we have at least one toss plus the number of additional tosses, lets call it $N'$. Note that $\E[N] = \E[N']$. Using similar logic we get:
$\E[N^2] = \E[(1 + N')^2]P(X=0) + \E[2 + N')^2]P(X=1) + \E[2]P(X=2)$.
$\E[N^2] = \E[(1 + N')^2]0.5 + \E[2 + N')^2]0.25 + \E[2]0.25$.
$\E[N^2] = 0.5+E[N]+0.5E[N^2]+ 1 + E[N] + 0.25E[N^2] + 1$.
$\E[N^2] = 2.5+2E[N]+0.75E[N^2]$.
$0.25\E[N^2] = 2.5+12$.
$\E[N^2] = 58$.
Thus $\Var[N] = 58 - 6^2 = 22$.
Seems a bit large to me, but I think that my steps are correct.