Find the probability that the lower face of the coin is a head 
A man possesses five coins, two of which are double-head (DH), one is double-tail (DT), and two are normal (N). He shuts his eyes, pick a coin at random, and tosses it. What is the probability that the lower face of the coin is a head. 
He open his eyes and sees that the coin is showing heads; what is the probability that the lower face is a head? He shuts his eyes again and tosses the coin again. What is the probability that the lower face is a head? He open his eyes and sees that the coin is show heads; what is the probability that the lower face is a head?
He discards this coin, picks another at random, and tosses it. What is the probability that it shows heads?

When the man tosses a coin at the first time, the only information we have are those five coins. Let $H_1$ denote as first toss, the probability that the lower face is a head at the first toss can be $$\mathbb{P}(H_1)=\mathbb{P}(H\vert DH)+\mathbb{P}(H\vert DT)+\mathbb{P}(H\vert N)=2/5+0+1/2\times 2/5=3/5$$ 
At his second toss $H_2$, we know that probability of first toss is $3/5$ which will effect the second toss, so $$\mathbb{P}(H_2\vert H_1)=\frac{\mathbb{P}(H_2\cap H_1)}{\mathbb{P}(H_1)}=\frac
{\mathbb{P}(DH)}{\mathbb{P}(H_1)}=(2/5)/(3/5)=2/3$$ His second toss will effect his third toss, I might get $\mathbb{P}(H_3\vert (H_2\vert H_1))$ which I don't how to calculate and don't know it is a legal expression or not. I get stuck for the remain questions.
 A: The law of total probability is: $$\begin{align}\mathsf P(H_1)~=~& \mathsf P(H_1\cap DH)+\mathsf P(H_1\cap DT)+\mathsf P(H_1\cap N) \\ ~=~& \mathsf P(H_1\mid DH)\mathsf P(DH)+\mathsf P(H_1\mid DT)\mathsf P(DT)+\mathsf P(H_1\mid N)\mathsf P(N) \\ =~& 1\cdot \tfrac 25 + 0\cdot\tfrac 1 5+\tfrac 1 2\cdot\tfrac 2 5 \\ =~& \frac 35\end{align}$$
So you accidentally arrived at the right value.

$\mathsf P(H_1\mid H_2) = \mathsf P(H_1\cap H_2)/\mathsf P(H_2) = \tfrac 25/\tfrac 35 = \tfrac 2 3$ as you had.

$\require{cancel}\color{red}{\xcancel{\color{black}{\mathsf P(H_3\mid(H_2\mid H_1))}}}$ is indeed not a valid construction.    There is only ever one divider between the event and the condition.
In any case, what you seek if $\mathsf P(H_3\mid H_2)$ : the probability that the lower face of the second toss is heads when given that the upper face of the first toss was heads.
Use similar reasoning as above.
$\mathsf P(H_3\mid H_2) ~=~ \dfrac{\mathsf P(H_3\cap H_2)}{\mathsf P(H_2)}~=~\dfrac{\mathsf P(H_2\cap H_3\cap DH)+0+\mathsf P(H_2\cap H_3\cap N)}{\mathsf P(H_2)}$

Now find $\mathsf P(H_3\mid H_2\cap H_4)$, the probability that the lower face of the second toss is heads given that the upper face of both tosses were each heads.
A: Let $H_L^i$/$H_U^i$ be the events that the lower/upper face is a head respectively on the $i$-th toss.
We want to compute the probability the lower face is a head on the second toss, given the upper face was a head on the first toss.
That is, we wish to compute $P(H_L^2|H_U^1)$. Using the definition of conditional probability, and then conditioning on $DH$, $N$ and $DT$:
$$
\begin{align*}
P(H_L^2|H_U^1)&=\frac{P(H_L^2\cap H_U^1)}{P(H_U^1)}\\\\
&=\frac{1}{P(H_U^1)}\left(P(H_L^2\cap H_U^1|DH)\cdot P(DH)+P(H_L^2\cap H_U^1|N)\cdot P(N)+P(H_L^2\cap H_U^1|DT)\cdot P(DT)\right)\\\\
&=\frac{1}{P(H_U^1)}\left(1\cdot\frac{2}{5}+\frac{1}{4}\cdot\frac{2}{5}+0\cdot\frac{1}{5}\right)\\\\
&=\frac{1}{2P(H_U^1)}
\end{align*}
$$
Note that $P(H_U^1)=1\cdot\frac{2}{5}+\frac{1}{2}\cdot\frac{2}{5}+0\times\frac{1}{5}=\frac{3}{5}$
Thus,
$$
P(H_L^2|H_U^1)=\frac{1}{2P(H_U^1)}=\frac{1}{2\cdot\frac{3}{5}}=\frac{5}{6}
$$
