What is the necessary volume of concentrated HCl to prepare a solution with a pH of 1.8?
Unfortunately no, you're not right. As mentioned in a now deleted comment by Bruno, your answer gives a $\mathrm{pH}$ of $-0.74$, which is not correct. You didn't give detailed steps of your calculations, and your first step doesn't seem like anything we should do.

We can solve this by making the calculations step by step. Let's start with the concentration of $\ce{H+}$, which will be the same as the concentration of $\ce{HCl}$, we will achieve:

$$\ce{[H+]} = 10^{-\mathrm{pH}} = \pu{0.01584893 mol//L}$$

Since we're going to prepare 5 liters, this means the amount of substance of $\ce{H+}$ should be

$$5 \times \pu{0.01584893 mol} = \pu{0.07924466 mol}$$

To transform this into amount of $\ce{HCl}$, we need the acid's molar mass ($\pu{36.46 g//mol}$) to convert from amount of substance to mass:

$$\require{cancel}%
g_{\ce{HCl}} 
= \mathrm{\frac{36.46\ g}{\cancel{mol}}} 
  \times 0.07924466\ \mathrm{\cancel{mol}} 
= 2.88926 \ \mathrm{g}$$

Now that we have the amount of substance of $\ce{HCl}$, we calculate the volume, which is where the density comes in:

$$V_{\mathrm{HCl}} 
= \mathrm{\frac{2.88926\ \cancel{g}}{1.179\ \cancel{g}\ mL^{-1}}} 
= 2.450602 \ \mathrm{mL}$$

Now the last step: Our stock solution isn't pure $\ce{HCl}$, it's diluted in water. So actually the volume we need is:

$$V_{\mathrm{HCl}} 
= 2.450602 \times \frac{100}{36} 
= \pu{6.807229 mL} 
\approx \pu{6.8 mL}$$

This might look that it is not much, but remember $\ce{HCl}$ is a very strong acid. For example, to prepare $\pu{1000 mL}$ of a $\pu{0.001 M}$ solution you only need $\pu{0.086 mL}$ of concentrated $\ce{HCl}$!

You can check this result by imputing the given density, weight percentage, final volume and $\ce{H+}$ concentration (first step) into Sigma-Aldrich's Molarity Calculator.