Evaluate the limit _x 3 (x2 - 9)/(x2 - 2x - 3) and determine if the function f(x) is continuous at x=2. If not, identify the type of discontinuity.

Since we have an indeterminate form, we need to simplify the expression.

The numerator is a difference of squares: $x^2 - 9 = (x-3)(x+3)$.
The denominator is a quadratic: $x^2 - 2x - 3 = (x-3)(x+1)$.

Since $x \to 3$, $x \neq 3$, so we can cancel the $(x-3)$ terms.

The limit is $\boxed*{\frac{3}{2}}*$.

If not, identify the type of discontinuity.

For a function to be continuous at $x=c$, three conditions must be met:

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

From the definition of the function, $f(2) = 3$. So, $f(2)$ is defined.

For $x \neq 2$, $f(x) = \frac{x^2 - 4}{x-2}$.

Factor the numerator: $x^2 - 4 = (x-2)(x+2)$.

Cancel the $(x-2)$ terms since $x \neq 2$.

Substitute $x=2$:

So, $\lim_{x \to 2} f(x) = 4$. The limit exists.

We have $f(2) = 3$ and $\lim_{x \to 2} f(x) = 4$.
Since $f(2) \neq \lim_{x \to 2} f(x)$, the function is not continuous at $x=2$.

Since the limit $\lim_{x \to 2} f(x)$ exists but is not equal to $f(2)$, this is a removable discontinuity. If we redefined $f(2)$ to be 4 instead of 3, the discontinuity would be removed.

The function is not continuous at $x=2$. It has a $\boxed{\text{removable discontinuity}}$.

For $g(x)$ to be continuous at $x=1$, the left-hand limit, the right-hand limit, and the function value at $x=1$ must all be equal.

Using the first part of the piecewise function for $x \le 1$:

For $x < 1$, $g(x) = 2x + k$.

For $x > 1$, $g(x) = 5 - x$.

For continuity, we must have:

Subtract 2 from both sides:

The value of $k$ that makes the function continuous at $x=1$ is $\boxed{2}$.