$$\begin{aligned} & \mathrm{P}\left(\mathrm{a}^2, \mathrm{a}+1\right) \\ & \mathrm{L}_1=3 \mathrm{x}-\mathrm{y}+1=0 \end{aligned}$$

Origin and $$\mathrm{P}$$ lies same side w.r.t. $$\mathrm{L}_1$$

$$\begin{aligned} & \Rightarrow \mathrm{L}_1(0) \cdot \mathrm{L}_1(\mathrm{P})>0 \\ & \therefore 3\left(\mathrm{a}^2\right)-(\mathrm{a}+1)+1>0 \end{aligned}$$

$$\begin{aligned} & \Rightarrow 3 \mathrm{a}^2-\mathrm{a}>0 \\ & \mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \quad \text{..... (1)} \end{aligned}$$

Let $$L_2: x+2 y-5=0$$

Origin and $$\mathrm{P}$$ lies same side w.r.t. $$\mathrm{L}_2$$

$$\begin{aligned} & \Rightarrow \mathrm{L}_2(0) \cdot \mathrm{L}_2(\mathrm{P})>0 \\ & \Rightarrow \mathrm{a}^2+2(\mathrm{a}+1)-5<0 \\ & \Rightarrow \mathrm{a}^2+2 \mathrm{a}-3<0 \\ & \Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0 \end{aligned}$$

$$\therefore \mathrm{a} \in(-3,1)\quad \text{..... (2)}$$

Intersection of (1) and (2)

$$\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)$$