A moment is defined as the cross-product of distance and force vectors:
$$\begin{align}
M &= r\times F \\
&= \det\left(\left|\begin{matrix}
i & j & k \\
r_x & r_y & r_z \\
F_x & F_y & F_z \\
\end{matrix}\right|\right) \\
&= (r_yF_z-r_zF_y)\hat{i} -(r_xF_z-r_zF_x)\hat{j} + (r_xF_y-r_yF_x)\hat{k}
\end{align}$$
In a 2D frame, $r_z = F_z = 0$, which simplifies this to $M = (r_xF_y-r_yF_x)\hat{k}$. So, a positive vertical force at a positive horizontal distance (to the right of the studied node) results in a positive moment. Meanwhile, a positive horizontal force at a positive vertical distance (above the studied node) results in a negative moment.
A simple way of thinking about it is that force components are positive to the right or upwards and distances are positive if a positive force would generate a positive (counter-clockwise) rotation.
In your last equation, you made the moment due to the distributed load along $\overline{AB}$ positive, when it should be negative. After all, a positive horizontal force at a positive vertical distance (above $A$) results in a negative moment.
It should, however, be:
$$\begin{align}
\sum M_A =& M_A \\
& - (3)\cdot(6\cdot20) \\
& + (4)\cdot(-60) \\
& + (14.5)\cdot(-7\cdot30) \\
& + (13)\cdot (F_{E,y})= 0 \\
\therefore M_A =& -1132.5\text {kN}
\end{align}$$
I personally hate writing equations like this, with signs in the middle of the multiplications, but it makes all the variables explicit. So, in this calculation:
- all the distances are positive (everything is either above or to the right of $A$)
- the horizontal distributed load is positive
- the vertical loads are negative
- $F_{E,y}$ is whatever it will be (positive, in our case)
- When calculating the moment due to vertical forces, the sign is positive
- When calculating the moment due to horizontal forces, the sign is negative