To calculate the pH of a $0.365\% (w/v)$ $HCl$ solution, we first determine the molarity of the solution. A $0.365\% (w/v)$ solution means there is $0.365 \text{ g}$ of $HCl$ in $100 \text{ ml}$ of solution. Therefore, in $1000 \text{ ml}$ ($1 \text{ L}$) of solution, the amount of $HCl$ is: $$ \begin{aligned} \text{Mass of } HCl &= \frac{0.365 \text{ g} \times 1000 \text{ ml}}{100 \text{ ml}} = 3.65 \text{ g} \end{aligned} $$ The molar mass of $HCl$ is $36.5 \text{ g/mol}$. Thus, the molarity ($M$) is: $$ \begin{aligned} \text{Molarity } [H^+] &= \frac{3.65 \text{ g}}{36.5 \text{ g/mol} \times 1 \text{ L}} = 0.1 \text{ M} \end{aligned} $$ Now, calculating the pH: $$ \begin{aligned} pH &= -\log[H^+] \\ &= -\log(0.1) \\ &= 1 \end{aligned} $$ Thus, the pH of the solution is $1$.