How do you determine the pH of a solution at different volumes?
With regard to your understanding of the correlation between concentration and volume, you should know that concentration $c$ is defined as
$$c=\frac nV$$
where $n$ is amount of substance and $V$ is the volume of the solution.

When you dilute a sample, you change the volume $V$, but you do not change the amount of substance $n$ in your sample. Therefore, if you know the initial concentration $c_1$ in the initial volume $V_1$
$$c_1=\frac n{V_1}$$
you can calculate the new concentration $c_2$ after dilution to the new volume $V_2$:
$$c_2=\frac n{V_2}$$
You can rearrange both equations to calculate the amount of substance $n$:
$$\begin{align}
c_1\cdot V_1&amp;=n\\[6pt]
c_2\cdot V_2&amp;=n
\end{align}$$
Since the amount of substance $n$ is constant, you get another useful equation:
$$c_1\cdot V_1=c_2\cdot V_2$$



In your exercise, after addition of all solutions, the new volume may be estimated as 
$$V_2\approx1\ \mathrm{ml}+0.05\ \mathrm{ml}+6.95\ \mathrm{ml}=8\ \mathrm{ml}$$

The initial concentration of the base (sodium acetate) is $c_{\text{base},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{base},1}=1\ \mathrm{ml}$.

Therefore, the new concentration of the base is

$$\begin{align}
c_{\text{base},1}\cdot V_{\text{base},1}&amp;=c_{\text{base},2}\cdot V_2\\[6pt]
c_{\text{base},2}&amp;=\frac{c_{\text{base},1}\cdot V_{\text{base},1}}{V_2}\\[6pt]
&amp;=\frac{0.10\ \mathrm{mol\ l^{-1}}\times1\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt]
&amp;=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$

The initial concentration of the acid (acetic acid) is $c_{\text{acid},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{acid},1}=0.05\ \mathrm{ml}$.

Therefore, the new concentration of the acid is

$$\begin{align}
c_{\text{acid},1}\cdot V_{\text{acid},1}&amp;=c_{\text{acid},2}\cdot V_2\\[6pt]
c_{\text{acid},2}&amp;=\frac{c_{\text{acid},1}\cdot V_{\text{acid},1}}{V_2}\\[6pt]
&amp;=\frac{0.10\ \mathrm{mol\ l^{-1}}\times0.05\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt]
&amp;=0.000625\ \mathrm{mol\ l^{-1}}
\end{align}$$

Finally, you can use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of the acetate buffer from the new concentrations $c_{\text{acid},2}$ and $c_{\text{base},2}$.