To determine if a function F(x) is continuous at a point x=c, three conditions must be met:

To determine if a function $F(x)$ is continuous at a point $x=c$, three conditions must be met:

1. $F(c)$ must be defined.
2. $\lim_{x \to c} F(x)$ must exist.
3. $\lim_{x \to c} F(x) = F(c)$.

1. Determine whether the function $F(x) = 3x^2 + 4x - 10$ is continuous at $x = 3$ or not.

Step 1: Evaluate $F(3)$.

$$F(3) = 3(3)^2 + 4(3) - 10$$ $$F(3) = 3(9) + 12 - 10$$ $$F(3) = 27 + 12 - 10$$ $$F(3) = 39 - 10$$ $$F(3) = 29$$

$F(3)$ is defined.

Step 2: Evaluate $\lim_{x \to 3} F(x)$.
Since $F(x)$ is a polynomial function, the limit as $x$ approaches 3 can be found by direct substitution.

$$\lim_{x \to 3} (3x^2 + 4x - 10) = 3(3)^2 + 4(3) - 10$$ $$\lim_{x \to 3} (3x^2 + 4x - 10) = 27 + 12 - 10$$ $$\lim_{x \to 3} (3x^2 + 4x - 10) = 29$$

The limit exists.

Step 3: Compare $F(3)$ and $\lim_{x \to 3} F(x)$.
Since $F(3) = 29$ and $\lim_{x \to 3} F(x) = 29$, we have $\lim_{x \to 3} F(x) = F(3)$.

Therefore, the function $F(x) = 3x^2 + 4x - 10$ is continuous at $x=3$.

$$\boxed{\text{Continuous}}$$

2. Is the function $F(x) = \frac{2x^2 + 3}{2x + 1}$ continuous at (i) $x = -1$ and (ii) $x = -\frac{1}{2}$?

A rational function is continuous everywhere except where its denominator is zero.
First, find the values of $x$ for which the denominator is zero:

$$2x + 1 = 0$$ $$2x = -1$$ $$x = -\frac{1}{2}$$

The function is undefined at $x = -\frac{1}{2}$.