Entropy of a dynamical system and source code length
I can try to answer your first question (but I'm not sure I understand exactly what you mean).

A source produces messages with average information $H$. You seem to have a binary source, with two messages $m_1$ and $m_2$.

Now you want to encode the messages to a sequence of 1s and 0s (subsequently called "bits"). This is called "source coding". Even though we know that, in principle, we should be able to encode $N$ messages using $NH$ bits (for large $N$), it is not always obvious how to do so. There are algorithms such as Huffman and Lempel-Ziv that can find good (even optimal) encodings. Note that all good coding algorithms encode messages in groups, not individually, and they approach $NH$ only when $N$ is large.

If your question is "what's the smallest number of bits that I need to encode $N$ messages from a source with entropy $H$", the answer is $NH$ (in theory), and somewhat more in practice, depending on the optimality of your encoding algorithm and the value of $N$.

Example 1: A source produces 8-bit messages $b_7b_6b_5b_4b_3b_2b_2b_1b_0$ uniformly at random, but $b_7$ is always 0. The entropy is 7 bits/message. Let's say we want to transmit 100 messages. Even though the source produced 800 bits, we only need 700 to transmit the messages.

Example 2: You flip a biased coin with $p(\text{heads})=0.3$ 1000 times. The entropy of the experiment is 0.88 bits/flip. To convey the result of the 1000 flips, you'll need on average 880 bits. Some particular results will need more bits, some will need less. To figure out how many bits you need for a particular sequence of flips you need to specify your encoder algorithm.

Example 3: You have a biased 3-sided die with probability distribution $p(1)=0.5$, $p(2)=0.25$, $p(3)=0.25$. The entropy is 1.5 bits/throw. It's pretty easy to come up with an encoder:

result  encoding
1    -&gt; 0
2    -&gt; 10
3    -&gt; 11


You should be able to convince yourself that, on average, this encoding results in 1.5 bits per throw. However, let's say that you're unlucky and you get 1000 '3' in a row. That particular sequence will need 2,000 bits to transmit. That's why the entropy can only predict the average sequence length, not the length you'll get for a particular set of messages.