You should use the synthesis equation of an impulse train with period $T$ (which is easy to derive):
$$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)=\sum_{k=-\infty}^{\infty}\frac{1}{T}e^{jk\frac{2\pi}{T} t}\tag{1}$$
That is: the Fourier coefficients for all terms is a constant ($\frac{1}{T}$).
Now assume that there are two impulses with different amplitudes $a$ and $b$ per period, or $$x(t)=a\delta(t)+b\delta(t+2),\ -4< t \le0$$ In such case we have
$$\begin{align}
x(t)&=\sum_{k=-\infty}^{\infty}a\cdot\delta(t-kT)+b\cdot\delta(t+\frac{T}{2}-kT)\\&=\sum_{k=-\infty}^{\infty}a\cdot\delta(t-4k)+b\cdot\delta(t+2-4k)\\
&=\sum_{k=-\infty}^{\infty}\frac{a}{4}e^{jk\frac{2\pi}{4} t}+\sum_{k=-\infty}^{\infty}\frac{b}{4}e^{jk\frac{2\pi}{4} (t+2)}\\
&=\sum_{k=-\infty}^{\infty}\frac{1}{4}e^{jk\frac{\pi}{2} t}\left(a+be^{jk\pi}\right)\\
&=\sum_{k=-\infty}^{\infty}\frac{1}{4}e^{jk\frac{\pi}{2} t}\left(a+b(-1)^k\right)\\
&=\begin{cases}
\displaystyle\sum_{k=-\infty}^{\infty}\frac{a+b}{4}e^{jk\frac{\pi}{2} t},& k\text{ even}\\[10pt]
\displaystyle\sum_{k=-\infty}^{\infty}\frac{a-b}{4}e^{jk\frac{\pi}{2} t},& k\text{ odd}
\end{cases}
\end{align}$$
Now refering again to $(1)$ and comparing it with the question, we should have
$$c_k=\begin{cases}
\frac{a+b}{4}=1\\[10pt]
\frac{a-b}{4}=2
\end{cases}$$
