Evaluate the limit of the expression inside the square root.

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^3$.

$$\lim_{x \to \infty} \frac{x^3-1}{4x^3-5x+3} = \lim_{x \to \infty} \frac{\frac{x^3}{x^3}-\frac{1}{x^3}}{\frac{4x^3}{x^3}-\frac{5x}{x^3}+\frac{3}{x^3}}$$ $$= \lim_{x \to \infty} \frac{1-\frac{1}{x^3}}{4-\frac{5}{x^2}+\frac{3}{x^3}}$$

As $x \to \infty$, terms like $\frac{1}{x^n}$ approach 0.

$$= \frac{1-0}{4-0+0} = \frac{1}{4}$$

$$\lim_{x \to \infty} \sqrt{\frac{x^3-1}{4x^3-5x+3}} = \sqrt{\lim_{x \to \infty} \frac{x^3-1}{4x^3-5x+3}} = \sqrt{\frac{1}{4}}$$ $$= \frac{1}{2}$$

The limit is $\boxed*{\frac{1}{2}}*$.

When $x \to -\infty$, $x$ is negative, so $\sqrt{x^2} = |x| = -x$. This means $x = -\sqrt{x^2}$.

$$\lim_{x \to -\infty} \frac{x}{\sqrt{x^2-1}} = \lim_{x \to -\infty} \frac{x}{\sqrt{x^2\left(1-\frac{1}{x^2}\right)}}$$ $$= \lim_{x \to -\infty} \frac{x}{|x|\sqrt{1-\frac{1}{x^2}}}$$

Substitute $|x| = -x$ for $x \to -\infty$:

$$= \lim_{x \to -\infty} \frac{x}{-x\sqrt{1-\frac{1}{x^2}}}$$

As $x \to -\infty$, $\frac{1}{x^2} \to 0$.

$$= \frac{-1}{\sqrt{1-0}} = \frac{-1}{1} = -1$$

The limit is $\boxed*{-1}*$.

$$\lim_{x \to \infty} (\sqrt{x^2+4x} - x) \cdot \frac{\sqrt{x^2+4x} + x}{\sqrt{x^2+4x} + x}$$ $$= \lim_{x \to \infty} \frac{(x^2+4x) - x^2}{\sqrt{x^2+4x} + x}$$ $$= \lim_{x \to \infty} \frac{4x}{\sqrt{x^2+4x} + x}$$

For $x \to \infty$, $\sqrt{x^2+4x} = \sqrt{x^2(1+\frac{4}{x})} = |x|\sqrt{1+\frac{4}{x}}$. Since $x \to \infty$, $x$ is positive, so $|x|=x$.

$$= \lim_{x \to \infty} \frac{\frac{4x}{x}}{\frac{x\sqrt{1+\frac{4}{x}}}{x} + \frac{x}{x}}$$ $$= \lim_{x \to \infty} \frac{4}{\sqrt{1+\frac{4}{x}} + 1}$$

As $x \to \infty$, $\frac{4}{x} \to 0$.

$$= \frac{4}{\sqrt{1+0} + 1} = \frac{4}{1+1} = \frac{4}{2} = 2$$

The limit is $\boxed{2}$.