The outcomes of the given experiment can be represented by the following tree diagram.
The sample space of the experiment is,
$S=\left\{\left(1,H\right),\left(1,T\right),\left(2,H\right),\left(2,T\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,H\right),\left(4,T\right),\left(5,H\right),\left(5,T\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\right\}$
Let $A$ be the event that the coin shows a tail and $B$ be the event that at least one die shows $3$.
$\therefore A=\left\{\left(1,T\right),\left(2,T\right),\left(4,T\right),\left(5,T\right)\right\}$
$B=\left\{\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(6,3\right)\right\}$
$\Rightarrow A\cap B=\varphi$
$\therefore P\left(A\cap B\right)=0$
Then, P(B) = P({3, 1}) + P({3, 2}) + P({3, 3}) + P({3, 4}) + P({3, 5}) + P({3, 6}) + P({6, 3})
\( =\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36} \\ =\frac{7}{36} \)
Probability of the event that the coin shows a tail, given that at least one die shows $3$, is given by $P\left(A|B\right)$.
Therefore, \(P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0}{\frac{7}{36}}=0\)

For independent events,
P(E∩F) = P(E) ⋅ P(F)
\(\Rightarrow 0.35=0.60 \times P(F) \Rightarrow P(F)=\frac{7}{12}=0.58\)

\( P(B / A)=\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4=P(A \cap B) \)
\(\Rightarrow P(A \cap B)=0.24 \)
\( P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

1/3

proof :
By formula of conditional probability, we know that
\( \mathrm{P}\left(\mathrm{E}_i \mid \mathrm{A}\right) =\frac{\mathrm{P}\left(\mathrm{A} \cap \mathrm{E}_i\right)}{\mathrm{P}(\mathrm{A})} \)
\(=\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{AlE} \mathrm{E}_i\right)}{\mathrm{P}(\mathrm{A})}(b y \) (by multiplication rule of probability)
\( =\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{AlE}_i\right)}{\sum_{j=1}^n \mathrm{P}\left(\mathrm{E}_j\right) \mathrm{P}\left(\mathrm{AlE}_j\right)} \)(by the result of theorem of total probability)

P(A) ⋅ P(B) = \(1\over2\)⋅\(1\over3\) = \(1\over6\) = P(A∩B) 
Yes, the events are independent.