Step 1: Recognise this as a stretched coupling case.
Combining $j_1=2$ with $j_2=\tfrac12$ can only give total angular momentum $j=\tfrac52$ (the stretched, maximum case) or $j=\tfrac32$. Since we want $j=\tfrac52$, which is $j_1+j_2$, there is a standard closed formula for the Clebsch-Gordan coefficients that mix a spin-$\tfrac12$ particle into a state of total $j=j_1+\tfrac12$, without needing to build up from the top state by hand.
Step 2: Quote the standard formula.
For coupling $j_1$ with a spin-$\tfrac12$ particle to reach $j=j_1+\tfrac12$ with projection $m$, the general result is:
\[ \left|j_1+\tfrac12,\,m\right\rangle = \sqrt{\frac{j_1+m+\tfrac12}{2j_1+1}}\;\Big|j_1,m-\tfrac12\Big\rangle\Big|\tfrac12,\tfrac12\Big\rangle \;+\; \sqrt{\frac{j_1-m+\tfrac12}{2j_1+1}}\;\Big|j_1,m+\tfrac12\Big\rangle\Big|\tfrac12,-\tfrac12\Big\rangle \]
This formula comes from the general Clebsch-Gordan tables for adding spin-$\tfrac12$ to any $j_1$, and it directly gives both mixing coefficients without any extra derivation.
Step 3: Plug in $j_1=2$ and $m=\tfrac32$.
Here $2j_1+1 = 5$. The first coefficient uses $m_1 = m-\tfrac12 = 1$, matching the term $|2,1\rangle|\tfrac12,\tfrac12\rangle$:
\[ \sqrt{\frac{j_1+m+\tfrac12}{2j_1+1}} = \sqrt{\frac{2+\tfrac32+\tfrac12}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt5} \]
The second coefficient uses $m_1 = m+\tfrac12 = 2$, matching the term $|2,2\rangle|\tfrac12,-\tfrac12\rangle$:
\[ \sqrt{\frac{j_1-m+\tfrac12}{2j_1+1}} = \sqrt{\frac{2-\tfrac32+\tfrac12}{5}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt5} \]
Step 4: Match to the question's labeling.
The question calls the coefficient of $|2,1\rangle|\tfrac12,\tfrac12\rangle$ as $c_1$ and the coefficient of $|2,2\rangle|\tfrac12,-\tfrac12\rangle$ as $c_2$, so directly $c_1 = \dfrac{2}{\sqrt5}$ and $c_2 = \dfrac{1}{\sqrt5}$.
Step 5: Check this against a quick consistency test.
A genuine quantum state must be normalised, meaning $c_1^2+c_2^2=1$. Here $\left(\dfrac{2}{\sqrt5}\right)^2 + \left(\dfrac{1}{\sqrt5}\right)^2 = \dfrac45+\dfrac15 = 1$, which checks out, confirming these are valid coefficients. Option (C), by contrast, also happens to satisfy $c_1^2+c_2^2=1$ but does not match the actual weighting of the two terms from the formula, so passing the normalisation check alone is not enough to pick it.
Final Answer:
The standard stretched-coupling formula gives $c_1=\dfrac{2}{\sqrt5}$ and $c_2=\dfrac{1}{\sqrt5}$, option (A).\[ \boxed{c_1=\tfrac{2}{\sqrt5},\ c_2=\tfrac{1}{\sqrt5}} \]