Given equations of lines are

$$\begin{gathered} y=4 x+5 \quad\text{.... (i)}\\ y=5-x \quad\text{.... (ii)}\\ \text{and}\quad 4 y=x+5 \quad \text{.... (iii)} \end{gathered}$$

On solving Eqs. (i) and (ii), we get

$$\begin{aligned} 4 x+5 & =5-x \\ \Rightarrow x & =0 \end{aligned}$$

On solving Eqs. (i) and (iii), we get

$$ \begin{aligned} 4(4 x+5) & =x+5 \\ \Rightarrow\quad 16 x+20 & =x+5 \\ \Rightarrow\quad 15 x & =-15 \\ \Rightarrow\quad x & =-1 \end{aligned}$$

$$\begin{aligned} &\text { On solving Eqs. (ii) and (iii), we get }\\ &\begin{array}{r} 4(5-x)=x+5 \\ \Rightarrow 20-4 x=x+5 \end{array} \end{aligned}$$

On solving Eqs. (i) and (iii), we get

$$\begin{aligned} &\text { On solving Eqs. (ii) and (iii), we get }\\ &\begin{array}{r} 4(5-x)=x+5 \\ \Rightarrow\quad 20-4 x=x+5 \end{array} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad x=3\\ & \therefore \text { Required area }=\int_{-1}^0(4 x+5) d x+\int_0^3(5-x) d x-\frac{1}{4} \int_{-1}^3(x+5) d x \\ &=\left[\frac{4 x^2}{2}+5 x\right]_{-1}^0+\left[5 x-\frac{x^2}{2}\right]_0^3-\frac{1}{4}\left[\frac{x^2}{2}+5 x\right]_{-1}^3 \\ &=[0-2+5]+\left[15-\frac{9}{2}-0\right]-\frac{1}{4}\left[\frac{9}{2}+15-\frac{1}{2}+5\right] \\ &=3+\frac{21}{2}-\frac{1}{4} \cdot 24 \\ &=-3+\frac{21}{2}=\frac{15}{2} \text { sq units } \end{aligned}$$

$$\begin{aligned} \text { Required area of shaded region } & =\int_0^{2 \pi} 2 \cos x d x \\ & =\int_0^{\pi / 2} 2 \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} 2 \cos x d x\right|+\int_{3 \pi / 2}^{2 \pi} 2 \cos x d x \end{aligned}$$

$$\begin{aligned} & =2[\sin x]_0^{\pi / 2}+\left|2(\sin x)_{\pi / 2}^{3 \pi / 2}\right|+2[\sin x]_{3 \pi / 2}^{2 \pi} \\ & =2+4+2=8 \text { sq units } \end{aligned}$$

We have, $y=1+|x+1|, x=-3, x=3$ and $y=0$

$\because\quad y=\left\{\begin{array}{cc}-x, & \text { if } x<-1 \\ x+2, & \text { if } x \geq-1\end{array}\right.$

$$\begin{aligned} \therefore \quad \text { Area of shaded region, } A & =\int_{-3}^{-1}-x d x+\int_{-1}^3(x+2) d x \\ & =-\left[\frac{x^2}{2}\right]_{-3}^{-1}+\left[\frac{x^2}{2}+2 x\right]_{-1}^3 \\ & =-\left[\frac{1}{2}-\frac{9}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right] \\ & =-[-4]+[8+4] \\ & =12+4=16 \text { sq units } \end{aligned}$$