Problem
Let $\cal X$ be the alphabet $\{A, B, C, E, F\}$.
Let $\{X_i\}$ be stochastic process with $\Pr\{X_i=x\}=p(x)$.
There is given that $p(A)=\displaystyle\frac12$, $p(B)=\displaystyle\frac14$, $p(C)=\displaystyle\frac18$, $p(D)\displaystyle\frac1{16}$, $p(E)=\displaystyle\frac1{32}$, $p(F)=\displaystyle\frac1{32}$.
Q1) Find the entropy of $X_1$.
A1) The entropy of $X_1$ is $$ H(X_1)=\sum_{x\in\cal{X}} p(x)\log_2\frac1{p(x)}=\sum_{k=1}^5\frac{k}{2^k}+\frac{5}{32}=\frac{31}{16}=1.9375 \text{ [bits]} $$
Q2) Find the most probable sequence.
A2) The most probable sequence is one that has all elements as $A$.
Q3) Is the most probable sequence in the typical set with $\varepsilon=0.1$.
A3) The typecal set is defined as $$ A_\varepsilon^{(n)}=\left\{(x_1, \ldots, x_n):\left\vert-\frac1n\log_2p(x_1,\ldots,x_n)-H(X)\right\vert\le\varepsilon\right\}. $$ The sequence whose all elements are $A$ has probability of $\Pr\{X_1=A,\ldots,X_n=A\}=2^{-n}$.
Since $\left\vert-\frac1n\log_2p(x_1,\ldots,x_n)-H(X)\right\vert=\left\vert1-1.9375\right\vert=0.9375\gt\varepsilon=0.1$, the most probable sequence is not in the typical set.
Q4) Find a shortest sequence in the typical set for a sufficiently small $\varepsilon\gt0$.
A4) I don't know.
Is the A1 to A3 correct? Moreover, please let me know how to solve Q4.
Firstly, you must assume that the stochastic process is i.i.d, with each sample distributed as $p(x)$ which is not explicitly stated.
A1 to A3 are correct. For solving Q4, let $f(x)=p(x) \log(1/p(x))$ and choose some positive $\varepsilon>0.$
Obviously the order of symbols chosen makes no difference. What you want is to choose nonnegative integers $N_A,N_B,N_C,N_E,N_F$ (i.e., a word with $N_A$ A's, $N_B$, B's, etc) such that $N=N_A+N_B+N_C+N_E+N_F$ is minimal and the sample entropy of such a word, namely $$ N(A) f(A) + N(B) f(B) + N(C) f(C) + N(E) f(E) + N(F) f(F) $$ satisfies $$ | N(A) f(A) + N(B) f(B) + N(C) f(C) + N(E) f(E) + N(F) f(F) - N (1.9375)| \leq N \varepsilon. $$ Those $f$ values, the probabilities being powers of two, are very special and your problem becomes one of elementary combinatorics how to take integer combinations of them to get close to your target.