One-One (Injective):

To prove that the function is one-one, we need to show that if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Let $f(x_1) = f(x_2)$. Then,

$\frac{5x_1 - 3}{4} = \frac{5x_2 - 3}{4}$

$5x_1 - 3 = 5x_2 - 3$

$5x_1 = 5x_2$

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f(x)$ is one-one.

Onto (Surjective):

To prove that the function is onto, we need to show that for every $y \in \mathbb{R}$, there exists an $x \in \mathbb{R}$ such that $f(x) = y$.

Let $y \in \mathbb{R}$. We want to find an $x$ such that $f(x) = y$.

$f(x) = \frac{5x - 3}{4} = y$

$5x - 3 = 4y$

$5x = 4y + 3$

$x = \frac{4y + 3}{5}$

Since $y \in \mathbb{R}$, $x = \frac{4y + 3}{5}$ is also a real number. Thus, for every $y \in \mathbb{R}$, there exists an $x = \frac{4y + 3}{5} \in \mathbb{R}$ such that $f(x) = y$. Therefore, the function $f(x)$ is onto.