**HINT**
According to these informations, $\textbf{P}(A) = 0.16$, $\textbf{P}(B) = 0.19$ and $\textbf{P}(A^{c}\cap B^{c}) = 1 - \textbf{P}(A\cup B)$, where $\textbf{P}(A\cup B) = \textbf{P}(A) + \textbf{P}(B) - \textbf{P}(A\cap B)$. Thus
(b) The answer is $\textbf{P}(A\cup B)$, which can be easily computed using the information above.
(c) The event in which we are interested is $(A\cap B^{c})\cup(A^{c}\cap B)$. Since these events are disjoint, \begin{align*} \textbf{P}(A\cap B^{c}) + \textbf{P}(A^{c}\cap B) & = \textbf{P}(A) - \textbf{P}(A\cap B) + \textbf{P}(B) - \textbf{P}(A\cap B)\\\ & = \textbf{P}(A) + \textbf{P}(B) - 2\textbf{P}(A\cap B) \end{align*}
(d) The event of interest is $\textbf{P}(B|A)$. Thus \begin{align*} \textbf{P}(B|A) = \frac{\textbf{P}(A\cap B)}{\textbf{P}(A)} = \ldots \end{align*}
(e) Analogously, we have $\textbf{P}(A|B)$, which equals to \begin{align*} \textbf{P}(A|B) = \frac{\textbf{P}(A\cap B)}{\textbf{P}(B)} = \ldots \end{align*}
Hope this helps.