Considering the image we observe the following: $$r_A (if $$0) and thus

$$\frac{K(+Q)}{r_B}=V(B) But we also know that the electric potential energy in a point $$P$$ is defined as $$V(P)=\frac{U}{q_0}$$, so if we consider a charge $$+q$$ where $$q>0$$ we have: $$U_B Does that mean that the "potential energy of $$+q$$" is lower if $$+q$$ is far from the charge $$+Q$$ and higher if it's close? Now, if we consider $$+Q$$ and $$-q$$ where $$Q>0$$ and $$q>0$$then we have: $$U_A Does that mean that the "potential energy of $$-q$$" is higher if $$-q$$ is far from the charge $$+Q$$ and lower if it's close? Now if we consider $$(-Q),Q>0$$ we have: $$\frac{K(-Q)}{r_A}=V(A) Again, if we consider a charge $$+q$$ where $$q>0$$ we have: $$U_A Does that mean that the "potential energy of $$+q$$" is higher if $$+q$$ is far from the charge $$-Q$$ and lower if it's close? And finally, if we consider a charge $$-q, q>0$$, we have: $$U_B Does that mean that the "potential energy of $$-q$$" is lower if $$-q$$ is far from the charge $$+Q$$ and higher if it's close?

Does that mean that the "potential energy of $$+q$$" is lower if +q is far from the charge $$+Q$$ and higher if it's close?

The potenial energy for a point is given by: $$V_p(r)=k_e\frac{qQ}{r}$$

where:

$$k_e=\frac{1}{4\pi \epsilon_0}$$

The value of $$V_p$$ does depends on:

1. the scalar values of $$q$$ and $$Q$$
2. if both charges are of equal sign then $$V_p$$ will be positive valued
3. if both charges are of opposite sign then $$V_p$$ will be negative valued.

The potential energy of the charge q at point A in the field is equal to the work required to move it to that point from an infinite distance* ($$U = -\int_{\infty}^{\mathbf{r}_A} q \mathbf{E} \cdot d \mathbf{r} = \frac{1}{4 \pi \epsilon_0} \frac{Q q}{r_A}$$). For like charges (+Q, +q or -Q, -q) this will be positive and decrease with increasing distance. For unlike charges (-Q, +q or +Q, -q) this will be negative and increase with increasing distance.

*The reference point implied by the formulae you cite.