We are given the function $f(x) = \sqrt{1 - \frac{x+13}{x^2+4x+3}}$ and asked to find its domain and discuss its continuity at $a=2$. We are also given the function $g(x) = \frac{\sqrt{2x+3} - 1}{\lfloor 4x^2 - 36 \rfloor}$ and asked to find its domain and discuss its continuity at $a = -\frac{3}{2}$. Here, $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$.

1. Problem Description

We are given the function $f(x) = \sqrt{1 - \frac{x+13}{x^2+4x+3}}$ and asked to find its domain and discuss its continuity at $a=2$.

We are also given the function $g(x) = \frac{\sqrt{2x+3} - 1}{\lfloor 4x^2 - 36 \rfloor}$ and asked to find its domain and discuss its continuity at $a = -\frac{3}{2}$. Here, $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$.

2. Solution Steps

1.1.a) Domain of $f(x)$:

For $f(x)$ to be defined, we need two conditions:

(1) The expression inside the square root must be non-negative: $1 - \frac{x+13}{x^2+4x+3} \geq 0$.

(2) The denominator of the fraction must be non-zero: $x^2+4x+3 \neq 0$.

First, consider the denominator: $x^2+4x+3 = (x+1)(x+3)$.

So, $x \neq -1$ and $x \neq -3$.

Next, consider the expression inside the square root:

$1 - \frac{x+13}{x^2+4x+3} \geq 0$

$\frac{x^2+4x+3 - (x+13)}{x^2+4x+3} \geq 0$

$\frac{x^2+3x-10}{x^2+4x+3} \geq 0$

$\frac{(x+5)(x-2)}{(x+1)(x+3)} \geq 0$

We can analyze the sign of this expression using a number line. The critical points are $x=-5, -3, -1, 2$.

- $x < -5$: All factors are negative, so the expression is positive.

- $-5 < x < -3$: $(x+5)$ is positive, the rest are negative, so the expression is negative.

- $-3 < x < -1$: $(x+5)$ and $(x+3)$ are positive, the rest are negative, so the expression is positive.

- $-1 < x < 2$: $(x+5)$, $(x+3)$ and $(x+1)$ are positive, $(x-2)$ is negative, so the expression is negative.

- $x > 2$: All factors are positive, so the expression is positive.

Thus, the solution to the inequality is $x \in (-\infty, -5] \cup (-3, -1) \cup [2, \infty)$.

So, the domain of $f(x)$ is $(-\infty, -5] \cup (-3, -1) \cup [2, \infty)$.

1.1.b) Continuity of $f(x)$ at $a=2$:

Since $2$ is in the domain of $f(x)$, we need to check if $f(2)$ is defined, if $\lim_{x \to 2} f(x)$ exists, and if $\lim_{x \to 2} f(x) = f(2)$.

$f(2) = \sqrt{1 - \frac{2+13}{2^2+4(2)+3}} = \sqrt{1 - \frac{15}{4+8+3}} = \sqrt{1 - \frac{15}{15}} = \sqrt{1-1} = 0$.

Since the expression is defined at $x=2$, and $f(2) = 0$, we need to evaluate $\lim_{x \to 2} f(x)$.

Since 2 is in the domain of $f(x)$ and $f(x)$ is constructed of continuous functions in its domain, we have $\lim_{x \to 2} f(x) = f(2) = 0$.

Therefore, $f(x)$ is continuous at $x=2$.

1.2.a) Domain of $g(x)$:

For $g(x)$ to be defined, we need two conditions:

(1) The expression inside the square root must be non-negative: $2x+3 \geq 0$, which implies $x \geq -\frac{3}{2}$.

(2) The denominator must be non-zero: $\lfloor 4x^2 - 36 \rfloor \neq 0$.

This means $4x^2 - 36$ cannot be in the interval $[0, 1)$, or $0 \leq 4x^2 - 36 < 1$.

This gives $36 \leq 4x^2 < 37$, so $9 \leq x^2 < \frac{37}{4} = 9.25$.

This implies $- \sqrt{9.25} < x \leq -3$ or $3 \leq x < \sqrt{9.25}$.

Thus $x$ cannot be in these intervals.

The domain of $g(x)$ must satisfy $x \geq -\frac{3}{2}$ and $x \notin [-\sqrt{9.25}, -3] \cup [3, \sqrt{9.25}]$.

Since $x \geq -\frac{3}{2} = -1.5$, we only need to consider the condition $x \notin [3, \sqrt{9.25}]$.

So, the domain of $g(x)$ is $[-1.5, 3) \cup (\sqrt{9.25}, \infty)$.

1.2.b) Continuity of $g(x)$ at $a = -\frac{3}{2}$:

First, we evaluate $g(-\frac{3}{2})$.

$g(-\frac{3}{2}) = \frac{\sqrt{2(-\frac{3}{2})+3} - 1}{\lfloor 4(-\frac{3}{2})^2 - 36 \rfloor} = \frac{\sqrt{-3+3}-1}{\lfloor 4(\frac{9}{4}) - 36 \rfloor} = \frac{0-1}{\lfloor 9-36 \rfloor} = \frac{-1}{\lfloor -27 \rfloor} = \frac{-1}{-27} = \frac{1}{27}$.

Now let's discuss the right-hand limit, since $x \geq -\frac{3}{2}$:

$\lim_{x \to -\frac{3}{2}^+} g(x) = \lim_{x \to -\frac{3}{2}^+} \frac{\sqrt{2x+3}-1}{\lfloor 4x^2 - 36 \rfloor}$.

As $x \to -\frac{3}{2}^+$, $\sqrt{2x+3}-1 \to 0$.

Also, as $x \to -\frac{3}{2}$, $4x^2 - 36 \to 4(\frac{9}{4})-36 = 9-36 = -27$.

Therefore, $\lfloor 4x^2 - 36 \rfloor \to -27$ as $x \to -\frac{3}{2}$.

So, $\lim_{x \to -\frac{3}{2}^+} g(x) = \frac{0-1}{-27} = \frac{1}{27}$.

Since $\lim_{x \to -\frac{3}{2}^+} g(x) = g(-\frac{3}{2}) = \frac{1}{27}$, the function $g(x)$ is right-continuous at $x=-\frac{3}{2}$.

3. Final Answer

1.1.a) Domain of $f(x)$: $(-\infty, -5] \cup (-3, -1) \cup [2, \infty)$

1.1.b) $f(x)$ is continuous at $x=2$.

1.2.a) Domain of $g(x)$: $[-1.5, 3) \cup (\sqrt{9.25}, \infty)$

1.2.b) $g(x)$ is right-continuous at $x=-\frac{3}{2}$.