Solutions

$$\begin{aligned} 4 - x & < 3- 2x \\ 4 + x & < 3 \\ x & < -1 \end{aligned}$$

$$\begin{aligned} 5 - x^2 & < 8 \\ 5 & < 8 + x^2 \\ -3 & < x^2 \end{aligned}$$

Notice that $\forall{x} \in \mathbb{R}, x^2 > 0$. Thus, any real number would satisfy this equation.

$$\begin{aligned} 5 - x^2 & < - 2 \\ 5 & < -2 + x^2 \\ 7 & < x^2 \\ \implies & x < -\sqrt{7} \quad \vee \quad x > \sqrt{7} \end{aligned}$$

$$\begin{aligned} (x - 1)(x - 3) > 0 & \implies (x - 1 > 0 \wedge x - 3 > 0) \\ & \quad \vee \; \; (x - 1 < 0 \wedge x - 3 < 0) \end{aligned}$$

This is so since:

$$\begin{aligned} \forall{x} \in \mathbb{R}, x > 0, \exists{y, z}: x = yz \implies [ y > 0 \wedge z > 0 ] \vee [ y < 0 \wedge z < 0] \end{aligned}$$