You'll be pleased to know that your answers are all fairly good! As far as I can see, you are right in saying that the force on the electron is not at all related to its velocity, and you have calculated it correctly. I'll explain why the force is at 90 degrees in a bit of a long-winded way, but hopefully it will make sense!
You will find that calculations like this are much easier when you use vectors, rather than work in degrees all the time. I don't know what kind of level you're at, or whether you have even come across vectors before, but judging by the way you've drawn the diagram it looks like you have a reasonable knowledge of them!
The full formula for calculating the electric field is as follows:
\underline{E}=\frac{q}{4\pi\epsilon_{0}|\underline{r}|^{2}} \hat{\underline{r}}
Where \hat{\underline{r}} is known as the 'unit vector'. This describes the direction in which the field points. Another thing to remember is the sign of q: positive charge = +q, negative = -q.
Calculating the unit vector is easy. We'll calculate it for your top charge first. \hat{\underline{r}} = \frac{\underline{r}}{|\underline{r}|} In other words, you want to divide the vector pointing in the direction of the field by its magnitude. So, at x = 20cm, \underline{r}=20x-15y (getting from your charge to that point), and |\underline{r}|=\sqrt{20^{2}+15^{2}}. After simplifying, \hat{\underline{r}}=\frac{4x-3y}{5}
I won't go through the calculation now, but the unit vector for the electric field caused by the lower charge is \hat{\underline{r}}=\frac{4x+3y}{5} Can you see why?
The beauty with sorting things out this way is that adding the electric field becomes easy, since it is just a case of addition: E_{NET}=\frac{q}{4\pi\epsilon_{0}|\underline{r}|^{2}} \frac{4x-3y}{5} + \frac{-q}{4\pi\epsilon_{0}|\underline{r}|^{2}} \frac{4x+3y}{5} = \frac{q}{4\pi\epsilon_{0}|\underline{r}|^{2}} (\frac{-6y}{5}) = -2071 y N/C In other words, we've just shown that the E field has a magnitude of 2071N/C (close enough to yours!) and is pointing straight down the y axis, which is what you also showed, except you described its direction with '270 degrees'.
To answer your initial confusion as to why the electron experiences a force 'at 90 degrees' as you put it, F = qE = -q * -2071 y = 2071q y. The minus sign flips the direction to the +ve y direction. This is really easy to do when you use the notation that I have, but a bit more tricky if you're working in degrees.
I really hope this has helped a little! Let me know if you want me to word things again or stop writing long-winded answers.
