(1) Consider the equilibrium of $q_1$: the only forces on it come from $q_2$ and $q_3$, both lying to the right. For these forces to cancel, one must repel $q_1$ (push left) and the other must attract it (pull right). So $q_1 q_2$ and $q_1 q_3$ have opposite signs --- equivalently, $q_2$ and $q_3$ have opposite signs. By symmetry (equilibrium of $q_3$), $q_1$ and $q_2$ have opposite signs. Combining: $q_1$ and $q_3$ share one sign, $q_2$ has the opposite sign. (The middle charge is opposite in sign to the two outer charges.)

(2) Let $d_{12}$ and $d_{23}$ denote the distances between adjacent pairs, so $d_{13} = d_{12} + d_{23}$.

Equilibrium of $q_2$ (forces from $q_1$ and $q_3$ on $q_2$ have equal magnitudes):

$$\dfrac{|q_1 q_2|}{d_{12}^2} = \dfrac{|q_2 q_3|}{d_{23}^2} \;\Longrightarrow\; \dfrac{d_{12}}{d_{23}} = \sqrt{\dfrac{|q_1|}{|q_3|}}.$$

Write $d_{12} = \sqrt{|q_1|}\, k$ and $d_{23} = \sqrt{|q_3|}\, k$ for some scaling constant $k$. Then

$$d_{13} = d_{12} + d_{23} = (\sqrt{|q_1|} + \sqrt{|q_3|})\, k.$$

Equilibrium of $q_1$ (forces from $q_2$ and $q_3$ on $q_1$ have equal magnitudes):

$$\dfrac{|q_2|}{d_{12}^2} = \dfrac{|q_3|}{d_{13}^2} \;\Longrightarrow\; \dfrac{d_{13}}{d_{12}} = \sqrt{\dfrac{|q_3|}{|q_2|}}.$$

Substituting:

$$\dfrac{(\sqrt{|q_1|} + \sqrt{|q_3|})\, k}{\sqrt{|q_1|}\, k} = \sqrt{\dfrac{|q_3|}{|q_2|}}.$$

Multiplying both sides by $\sqrt{|q_1 q_2|}$:

$$\sqrt{|q_1 q_2|} + \sqrt{|q_2 q_3|} = \sqrt{|q_1 q_3|},$$

which rearranges to $\sqrt{|q_1 q_2|} + \sqrt{|q_2 q_3|} - \sqrt{|q_1 q_3|} = 0$. $\blacksquare$