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
You need to solve the following equation.
$\mathbb{E}(X) = \displaystyle 1 + \frac{2}{3} (1 + \frac{3}{4} \mathbb{E}(X))$
Explanation: Say the initial state is $I$. You get $H$ with probability $\frac{1}{3}$ in the first toss or continue with probability $\frac{2}{3}$ to the second coin. We get $H$ on the second coin with probability $\frac{1}{4}$ or we get back to initial state $I$ with probability $\frac{3}{4}$.