Alternately, you could fully build the tree. What is the probability that you picked a fair coin? What is the probability that it shows heads five times in a row? The unfair coin? Five times in a row?
And don't forget conditional probability.
For expectation - yes due to linearity of expectation. As for variance, you can do the same since the each toss (or each event) are independent. Note that it does not matter whether you toss each coin 'one at a time' or the order you do it in due to independence, of course.
Formally, if you let $X_1$ and $X_2$ be the indicator variables for each toss of the first coin ($1$ for heads, $0$ for tails) and the same with $Y_1$ and $Y_2$ for the tosses of the second coin, we have:
$$N=X_1+X_2+Y_1+Y_2$$
Then
$$E(N)=E(X_1+X_2+Y_1+Y_2)$$
$$=E(X_1)+E(X_2)+E(Y_1)+E(Y_2)$$
due to linearity of expectation.
Also, as $X_1,X_2,Y_1,Y_2$ are independent events, we have
$$Var(N)=Var(X_1+X_2+Y_1+Y_2)$$
$$=Var(X_1)+Var(X_2)+Var(Y_1)+Var(Y_2)$$
The general rules being used here are:
$$E(aX+bY)=aE(X)+bE(Y)\tag{linearity of expectation}$$
and if $X$ and $Y$ are independent,
$$Var(aX+bY)=a^2Var(X)+b^2Var(Y)$$
Best Answer
Lat $A$ be one of the fair coins. Let $p$ be the probability that the number of "heads" among the rest (i.e., the coins $\ne A$) is even. Then the probability for a total number of even "heads" is $$P(A\text{ tails})P(\text{rest even})+P(A\text{ heads})P(\text{rest odd})=\frac12p+\frac12(1-p)=\frac12.$$