Spectroscopic Determination of an Equilibrium Constant
You must first calculate dilution factors. Assuming the volumes are additive, we can calculate dilution using $M_1V_1=M_2V_2$ equation. Also, assuming $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated, we can say:

$$\ce{[Fe(NO3)3] = [Fe^3+]} \text{ and } \ce{[KSCN] = [SCN-]}$$

Thus, initial $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ are $0.20$ and $\pu{2.0 \times 10^{-4} M}$, respectively.

In test tube one: $\pu{5.0 mL}$ of $\pu{0.20 M} \; \ce{Fe(NO3)3}$ and $\pu{5.0 mL}$ of $\pu{2.0 × 10^{−4} M} \; \ce{KSCN}$ are added together, thus dilution factor is 1:2. Therefore, using $M_1V_1=M_2V_2$ equation you can calculate initial concentrations of $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ after the addition (but before they react with each other) as follows:

$$\ce{[Fe^3+]} = \pu{\frac{0.2 \times 5.0}{10.0} M} = \pu{0.10 M} \text{ and } \ce{[SCN-]} = \pu{\frac{2.0 × 10^{−4}  \times 5.0}{10.0} M} = \pu{1.0 × 10^{−4} M}$$

As a matter of fact, since the original solution of $\ce{KSCN}$ did not undergo any dilution maneuver before final 1:1 dilution, $\ce{[SCN-]}$ of all 6 test tubes are same, and its value is $\pu{1.0 × 10^{−4} M}$.

On the other hand, the original solution of $\ce{Fe(NO3)3}$ has undergone serial dilution of $\pu{10 mL}$ to $\pu{25 mL}$ (hence dilution factor is $10/25$ or $2/5$). Therefore, if original concentration of each solution is $x$ before add together in test tubes, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$ is: 

$$\ce{[Fe^3+]}_a = \frac{x \times 2.0}{5.0} \: \pu{M}$$

Thus, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$:

$$\ce{[Fe^3+]}_2 = \pu{\frac{0.20 \times 2.0}{5.0} M}= \pu{0.08 M}, \, \ce{[Fe^3+]}_3 = \pu{\frac{0.08 \times 2.0}{5.0} M}= \pu{0.032 M}, \,  \ce{[Fe^3+]}_4 = \pu{\frac{0.032 \times 2.0}{5.0} M}= \pu{0.0128 M}, \,  \ce{[Fe^3+]}_5 = \pu{\frac{0.0128 \times 2.0}{5.0} M}= \pu{0.00512 M}, \,  \ce{[Fe^3+]}_6 = \pu{\frac{0.00512 \times 2.0}{5.0} M}= \pu{0.002048 M}$$

Finally, those concentrations were further diluted in 1:1 ratio in test tubes, and therefore, initial $\ce{[Fe^3+]}$ are half of those $\ce{[Fe^3+]}_a$ values. I tabulated all of them accordingly in following table, as answer to your question 1:

$$
\begin{array}{c|cc}
\text{Test Tube #} &amp;\text{Initial $\ce{[Fe^3+]}$} &amp;\text{Initial $\ce{[SCN-]}$} \\\hline
1 &amp; \pu{0.10 M} &amp; \pu{1.0 × 10^{−4} M}\\
2 &amp; \pu{0.04 M} &amp; \pu{1.0 × 10^{−4} M}\\
3 &amp; \pu{0.016 M} &amp; \pu{1.0 × 10^{−4} M}\\
4 &amp; \pu{0.0064 M} &amp; \pu{1.0 × 10^{−4} M}\\
5 &amp; \pu{0.00256 M} &amp; \pu{1.0 × 10^{−4} M}\\
6 &amp; \pu{0.001024 M} &amp; \pu{1.0 × 10^{−4} M}\\\hline
\end{array}
$$


There are 7 different values in the table. But you have given one
relevant answer, and it is also incorrect.
Find the answer for $\ce{[SCN-]}$ in test tube 4 from the table, which is
identical in all 6 test tubes.
The concern equilibrium is: $$\ce{Fe^3+ + SCN- &lt;=&gt; FeSCN^2+}$$ Reaction mole ratio of $\ce{[Fe^3+]}$ : $\ce{[SCN-]}$ : $\ce{[FeSCN^2+]}$ is 1 : 1 : 1. If equilibrium concentration of $\ce{[FeSCN^2+]}$ in test tube 2 is $\ce{ 8.3×10^{−5} M}$, the equilibrium concentration of $\ce{[Fe^3+]}$ in test tube 2 is $\ce{ (0.04 - 8.3×10^{−5}) M}$ and that of $\ce{[SCN-]}$ in test tube 2 is $\ce{ (1.0×10^{−4} - 8.3×10^{−5}) M}$. Now, you may able to calculate equilibrium constent for the reaction using following equation:


$$K_\mathrm{eq} =  \frac{\ce{[FeSCN^2+]}}{\ce{[Fe^3+]}\ce{[SCN-]}}$$