Can I get some pointers on what to do in calculating the hardness of water?
The hardness of water is defined as the sum of the dissolved $\ce{Ca^2+}$ and $\ce{Mg^2+}$ species and can be reported in different ways (moles/L, g/L, and so on). The one I have seen most commonly is mg/L, so that is the calculation presented below.

You need to determine how many milligrams of $\ce{Ca^2+}$ and $\ce{Mg^2+}$ are present and then divide that number by the solution volume (here, 10 L).

For $\ce{Ca^2+}$:

$$1.64\;\mathrm{g}\;\ce{Ca(NO3)2}\cdot{1\;\mathrm{mol}\;\ce{Ca^2+}\over 164.09\;\mathrm{g}}\cdot{40078\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Ca^2+}}
 = 400\;\mathrm{mg}$$

For $\ce{Mg^2+}$:

$$0.94\;\mathrm{g}\;\ce{Mg(Cl)2}\cdot{1\;\mathrm{mol}\;\ce{Mg^2+}\over 95.2\;\mathrm{g}}\cdot{24305\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Mg^2+}}
= 240\;\mathrm{mg}$$

Thus, you have 640 mg $\ce{Ca^2+}$ and $\ce{Mg^2+}$ in 10 L of solution so the hardness is:

$${640\;\mathrm{mg}\over 10\;\mathrm{L}} = 64\;\mathrm{mg}\cdot\mathrm{L}^{-1}$$