Buffers calculating amounts of acid & base
I'm going to address only the first part since that was in the original question. 

You were given a buffer of pH 4.5 and the pKa of the acid used is 4.79. You worked out the volumes of the acid to be 9.7 mL and the volume of base to be 5.3 mL. You are now asked to get the amount of acid and the amount of base (i.e. the number of moles of each in the buffer).

$n_a$ means the number of moles of the acid.

This is equal to the concentration of the acid multiplied by the volume of the acid, namely:

$n_a = [\ce{HA}].V_a$

Since we know that the concentration of the acid, $[\ce{HA}]$ is 0.1 M and the volume of acid used is 9.7 mL, the number of moles of acid becomes:

$n_a = 0.1\ \text{mol/L}\times 9.7\ \text{mL} \times\dfrac{1\ \text{L}}{1000\ \text{mL}}$

The $\dfrac{1\ \text{L}}{1000\ \text{mL}}$ is only to convert mL to L. Do the same for the number of moles of the conjugate base.

For the second part, you need to realise that the pH you get that way will be the combined pH of the buffer and the acid. You will probably have to use the assumption that 1 drop = 1 mL, unless the question or your tutor already mentioned how to consider the volume of the added acid.