This problem can be solved by the standard tree technique. The answers are $\boxed{\frac{1}{2}}$, $\boxed{\frac{5}{12}}$, and $\boxed{\frac{7}{72}}$, respectively.

• The bug can travel from $A$ to $E$ in $2$ steps along two paths:

Therefore, the probability is $$\frac{1}{2}\times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} =\frac{1}{2}$$

The bug can move from $A$ to $F$ in $3$ steps along three paths.

Therefore, the probability is $$\frac{1}{2}\times \frac{1}{2}\times 1 + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{3}+ \frac{1}{2}\times \frac{1}{2}\times \frac{1}{3}=\frac{5}{12} $$

• Three paths can lead the bug from $A$ to $G$ in $4$ steps.

Therefore, the probability is $$\frac{1}{2}\times \frac{1}{2}\times \frac{1}{3}\times \frac{1}{3} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{3}\times \frac{1}{2}+ \frac{1}{2}\times \frac{1}{2}\times \frac{1}{3}\times \frac{1}{3}=\frac{7}{72}$$