I think you have a good hang of the concept. However, things can always be written better.
For example, the sample space for three independent dice, rather than the suggestive $\{(1,1,1),...,(6,6,6)\}$(which is correct, so credit for that) can be written succinctly as $\Xi \times \Xi \times \Xi$, where $\Xi = \{1,2,3,4,5,6\}$. This manages to express every element in the sample space crisply, since we know what elements of cartesian products look like.
The sigma field is a $\sigma$-algebra of subsets of $\Omega$. That is, it is a set of subsets of $\Omega$, which is closed under infinite union and complement.
Ideally, the sigma field corresponding to a probability space, is the set of events which can be "measured" relative to the experiment being performed i.e. it is possible to assign a probability to this event, with respect to the experiment being performed. What this specifically means, is that based on your experiment, your $\sigma$-field can possibly be a wise choice.
In our case, we have something nice : every subset of $\Omega$ can be "measured" since $\Omega$ is a finite set (it has $6^3 = 216$ elements) hence the obvious candidate, namely the cardinality of a set can serve as its measure(note : this choice is not unique! You can come up with many such $\mathcal F$).
Therefore, $\mathcal F = \mathcal P(\Omega)$, where $\mathcal P(S)$ for a given set $S$ is the power set of $S$, or the set of all subsets of $S$. This is logical since this contains every subset of $\Omega$, and is obviously a $\sigma$-algebra.
Now, $P(A)$, for $A \subset \mathcal F$(equivalently, $A$ any subset of $\Omega$) is, for the reasons I mentioned above, simply the ratio between its cardinality, and the cardinality of $\Omega$. Therefore, $P(A) = \frac{|A|}{216}$. That is exactly what you wrote, except well, division by sets isn't quite defined.
So there you have it, an answer, along with what you've done right and wrong.
NOTE : $\mathcal P$ for the power set, and $P$ for the probability of a set is my notation here. I still think the letter $P$ is used for both, but it's causing confusion here, hence the change.
So let's look at $P(\mathrm{sum}=S \mid \mathrm{die\ biased\ toward\ } k)$.
Each roll of the biased die is equivalent to a process where we first choose between case $A$ with probability $\frac{6}{5} \epsilon$ and case $B$ with probability $1 - \frac{6}{5} \epsilon$. In case $A$, the result of the process is $k$. In case $B$, we'll roll a fair die and use that as the result. The net probability works out to the stated $\frac{1}{6} + \epsilon$ for result $k$ and $\frac{1}{6} - \frac{1}{5} \epsilon$ for any result other than $k$. (I came up with the probabilities for $A$ and $B$ by requiring $P(B) \frac{1}{6}$ must be the probability the result is a particular number other than $k$.)
Pretending the biased die value instead comes from this process, let $M$ be the number of times case $B$ happens, so case $A$ happens $6-M$ times. The probability that case $B$ happens $m$ times is
$$ P(M=m) = {6 \choose m} \left(\frac{6}{5} \epsilon\right)^{6-m} \left(1 - \frac{6}{5} \epsilon\right)^m $$
If case $B$ happens $m$ times, then we have $6-m$ results of exactly $k$, and $m$ results equally likely to be any number from $1$ to $6$. If all $6$ add up to $S$, then the case $B$ values must add up to $S - (6-m)k$. The number of ways this could happen is $N_m^{S-(6-m)k}$. ($N_r^s$ is zero for any impossible combinations of $r$ and $s$, or if $r$ or $s$ is negative. But $N_0^0 = 1$.) So
$$ P(\mathrm{sum} = S \mid (\mathrm{die\ biased\ toward\ } k) \land (M=m)) =
\frac{N_m^{S-(6-m)k}}{6^m} $$
$$ \begin{align*}
P((M=m) \land (\mathrm{sum} = S) \mid \mathrm{die\ biased\ toward\ } k) &= {6 \choose m} \left(\frac{6}{5} \epsilon\right)^{6-m} \left(1 - \frac{6}{5} \epsilon\right)^m \frac{N_m^{S-(6-m)k}}{6^m} \\
&= {6 \choose m} \left( \frac{6}{5} \epsilon \right)^{6-m} \left(\frac{1}{6} - \frac{\epsilon}{5} \right)^m N_m^{S-(6-m)k}
\end{align*} $$
Combine all possible values of $m$ to get:
$$ P(\mathrm{sum} = S \mid \mathrm{die\ biased\ toward\ } k) = \sum_{m=0}^6 {6 \choose m} \left( \frac{6}{5} \epsilon \right)^{6-m} \left(\frac{1}{6} - \frac{\epsilon}{5} \right)^m N_m^{S-(6-m)k} $$
Then as you already described, you can combine all possible values of $k$ to get:
$$ P(\mathrm{sum = S} \mid \mathrm{unfair\ bag}) =
\frac{1}{6} \sum_{k=1}^6 \sum_{m=0}^6 {6 \choose m} \left( \frac{6}{5} \epsilon \right)^{6-m} \left(\frac{1}{6} - \frac{\epsilon}{5} \right)^m N_m^{S-(6-m)k} $$
Best Answer
Your answer to 1 is correct.
Your answer to 2 $A:$ the die rolled is a loaded $\lnot A$: not A or the die rolled is fair $B$ = the sum of 3 values is 4.
$P(B|A) = 3(\frac {2}{6})(\frac {2}{6})(\frac {1}{6})=\frac {12}{216}$
The only way to do it is if you roll $2$ two's and a one, and the one could show up on any of $3$ rolls.
$P(B|\lnot A) = 3(\frac {1}{6})(\frac {1}{6})(\frac {1}{6})=\frac {3}{216}$
$P(B) = P(A)P(B|A) + P(\lnot A)P(B|\lnot A) = (\frac 12)\frac {12}{216} + (\frac 12)(\frac {3}{216})$
And that gives you all of the necessary components to plug into Bayes law.
$\frac {12}{15} = \frac 45$