Let $T$ be the time spent in the mine. Conditioning on the first door the miner chooses, we get
$$ \mathbb{E}[T]=\frac{1}{3}\cdot3+\frac{1}{3}(5+\mathbb{E}[T])+\frac{1}{3}(7+\mathbb{E}[T])$$
so
$$ \mathbb{E}[T]=5+\frac{2}{3}\mathbb{E}[T].$$
If $\mathbb{E}[T]$ is finite, then we can conclude that $\mathbb{E}[T]=15$.
To see that $\mathbb{E}[T]$ is finite, let $X$ be the number of times the miner chooses a door. Then $ \mathbb{P}(X\geq n)=(\frac{2}{3})^{n-1}$ for $n=1,2,3,\dots$, hence
$$ \mathbb{E}[X]=\sum_{n=1}^{\infty}\mathbb{P}(X\geq n)=\sum_{n=1}^{\infty}\Big(\frac{2}{3}\Big)^{n-1}<\infty$$
And since $T\leq 7(X-1)+3$, we see that $\mathbb{E}[T]<\infty$ as well.
I'll renumber the doors $0$ to $6$ to simplify the calculations.
For the first question, we can set up a recurrence for the expected number $a_n$ of moves the mouse will make starting at door $n$:
$$
a_n=1+(1-p)a_{n-1}+pa_{n+1}\;,
$$
with $p=\frac13$. A particular solution of the inhomogeneous equation is $a_n=3n$, and the homogeneous equation can be solved using the ansatz $a_n=\lambda^n$, leading to the charateristic equation
$$
p\lambda^2-\lambda+1-p=0
$$
with solutions $\lambda=1$ and $\lambda=\frac1p-1=2$. Altogether,
$$
a_n=c_1+c_22^n+3n\;.
$$
The boundary values are $a_0=a_6=0$, which yields
\begin{eqnarray*}
c_1+c_2&=&0\;,\\
c_1+64c_2+18&=&0\;,
\end{eqnarray*}
with solution $c_1=-c_2=\frac27$. Thus the life expectancy in the middle is
$$
a_3=\frac27-\frac27\cdot2^3+3\cdot3=7\;.
$$
For the second question, you can group the $100$ steps into $50$ pairs, reducing the process to doors $1$, $3$ and $5$. The transition matrix for each pair of steps is
$$
\frac19\pmatrix{2&4&0\\1&4&4\\0&1&2}\;,
$$
which conveniently happens to have a nice eigensystem. The initial state decomposes as
$$
\pmatrix{0\\1\\0}=\frac16\left(\pmatrix{4\\4\\1}-\pmatrix{4\\-2\\1}\right)\;,
$$
where the first vector is an eigenvector with eigenvalue $\frac23$ and the second vector is an eigenvector with eigenvalue $0$. Thus after $50$ pairs of steps the distribution on doors $1$, $3$ and $5$ is
$$
\frac16\left(\frac23\right)^{50}\pmatrix{4\\4\\1}\;,
$$
and the sum of these probabilities is
$$
\left(\frac23\right)^{49}\approx2.4\cdot10^{-9}\;,
$$
so your guess had the right order of magnitude.
We can also use this eigenanalysis to derive the life expectancy another way. After the first pair of steps, the distribution is
$$
\frac19\pmatrix{4\\4\\1}\;.
$$
From then on, the mouse gets eaten with probability $\frac13$ in each pair of steps, so the expected number of pairs after the first one is $3$. Since death occurs on the first step of a pair, that translates to $7$ steps.
Best Answer
Consider the fate of the mouse after its first cycle. The outcome of the first cycle is either $L$ (went left and came back) with probability $\frac12$, $O$ (went right and exited) with probability $\frac16$, or $R$ (went right and came back) with probability $\frac13$. Thus, $$X=(3+X')\mathbf 1_L+2\mathbf 1_O+(5+X')\mathbf 1_R,$$ where $X'$ is distributed like $X$ and independent of $(L,O,R)$. Taking expectations yields $$E(X)=(3+E(X))P(L)+2P(O)+(5+E(X))P(R),$$ from which, since $P(L)=3P(O)$ and $P(R)=2P(O)$, one deduces that $$E(X)=\frac{3P(L)+2P(O)+5P(R)}{P(O)}=3\cdot3+2+5\cdot2=21.$$ Likewise, $$X^2=(3+X')^2\mathbf 1_L+4\mathbf 1_O+(5+X')^2\mathbf 1_R,$$ hence $$E(X^2)=E((3+X)^2)P(L)+4P(O)+E((5+X)^2)P(R),$$ from which... you might wish to pursue.