Solution:

Determining the sample space of the given experiment is $\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \Pi\}\}$

(i) Given that, E: tail appears on one coin

And F: one coin shows head

Sample space of events,

$\Rightarrow \mathrm{E}=\{\mathrm{HT}, \mathrm{TH}\}$ and $\mathrm{F}=\{\mathrm{HT}, \mathrm{TH}\}$

$\text{E}\cap \text{F}=\{\text{HT},\text{TH}\}$

The value of probability of events,  $P(E)=\frac{2}{4}=\frac{1}{2}, P(F)=\frac{2}{4}=\frac{1}{2}, P(E \cap F)=\frac{2}{4}=\frac{1}{2}$

Now, we know that conditional probability is given by relation:

$P(E \mid F)=\frac{P(E \cap F)}{P(F)}$

Substituting the values, we get

$\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{1 / 2}{1 / 2}$

$\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=1$

(ii) Given, E: no tail appears

And F: no head appears

Sample space of events,

$\Rightarrow \mathrm{E}=\{\mathrm{HH}\}$ and $\mathrm{F}=\{T \mathrm{~T}\}$

So common sample space,

$\Rightarrow E \cap F=\varphi$

 The value of probability of events,

 $P(E)=\frac{1}{4}, P(F)=\frac{1}{4}, P(E \cap F)=\frac{0}{4}=0$

Now, we know that conditional probability relation is,

$P(E \mid F)=\frac{P(E \cap F)}{P(F)}$

Substituting the values, we get

$$\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{0}{1 / 4}$$

$\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=0$