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Solution:-Given that,The cummulative distribution function of a continuous vandom variable x is,F(x)=\left\{\begin{array}{cl}0 ; & \text { for } x<-2 \\0.25 x+0.5 ; & \text { For }-2 \leq x<1 \\0.5 x+0.25 ; & \text { for } 1 \leq x<1.5 \\1 ; & \text { for } 1.5 \leq x\end{array}\right.i)\begin{array}{l}\text { For } \quad x<-2 \\f(x)=\frac{d}{d x}(F(x)) \\f(x)=\frac{d}{d x}(x) \\f(x)=0\end{array}ii) For -2 \leq x<1\begin{array}{l}f(x)=\frac{d}{d x}(0.25 x+0.5) \\f(x)=0.25\end{array}iii) for t \leq x<1.5\begin{array}{l}f(x)=\frac{d}{d x}(0.5 x+0.25) \\f(x)=0.5\end{array}iv) for 1.5 \leq x\begin{array}{l}f(x)=\frac{d}{d x} *(1) \\f(x)=0\end{array}Therefore p.d.f of x isf(x)=\left\{\begin{array}{cc}0.25 & \text { for }-2 \leq x<1 \\0.5 ; & \text { for } 1 \leq x<1.5 \\0 & \text { otherwise. }\end{array}\right.b)\begin{array}{l}P(x>1.2)=1-P(x \leqslant 1.2) \\P(x>1.2)=1-F(1.2) \\P(x>1.2)=1-(0.5 \times 1.2+0.25) \\P(x>1.2)=1-(0.85) \\P(x>1.2)=0.15\end{array}c)\begin{array}{l}P(0<x<1.25)=F(1.25)-F(0) \\P(0<x<1.25)=(0.5 \times 1.25+0.25)-(0.25 \times 0+0.5) \\P(0<x<1.25)=0.875-0.5 \\P(0<x<1.25)=0.375\end{array}d)\begin{aligned}P(0<x<1.25) & =\int_{0}^{1} 0.25 d x+\int_{1}^{125} 0.5 d x \\& =[0.25, x]_{0}^{1}+[0.5 \times x]_{1}^{1.25}\end{aligned}Date\begin{array}{l}P(0<x<1.25)=[0.25-0]+[0.5 \times 1.25-0.5 \times 1]] \\P(0<x<1.25)=0.25+0.625-0.5 \\P(0<x<1.25)=0.375\end{array}e) i) For any x such that, x<-2\begin{array}{l}F(x)=\int_{-\infty}^{x} 0 d t \\F(x)=0\end{array}ii) For ary x such that, -2 \leq x<1\begin{array}{l}F(x)=\int_{-\infty}^{-2} 0 d t+\int_{2}^{x} 0.25 d t \\F(x)=0+[0.25 x t]_{-2}^{x} \\F(x)=0.25 \times(x)-0.25 \times(-2) \\F(x)=0.25 . x+0.5\end{array}iii) For any x such that, 1 \leqslant x<1.5\begin{array}{l}F(x)=\int_{-\infty}^{-2} 0 d t+\int_{-2}^{1} 0.25 d t+\int_{1}^{x} 0.5 d t \\F(x)=0+\left[0.257 t 1+[0.5 t]_{1}^{x}\right. \\F(x)=0+[0.25+0.5]+[0.5 \neq-0.5]\end{array}\begin{array}{l}F(x)=0.75+0.5 \\F(x)=0.75-0.5+0.5 x \\F(x)=0.5 x+0.25\end{array}iv) For any x such that, 1.5 \leq x\begin{array}{l}F(x)=\int_{-\infty}^{-2} 0 d t+\int_{-2}^{1} 0.25 d t+\int_{1}^{1.5} 0.5 d t+\int_{1}^{1.5} 0 d t \\F(x)=0+[0.25 t]_{-2}^{1}+[0.5 \cdot t]_{1}^{1.5}+0 \\F(x)=(0.25 \times 1-0.25 \times(-2))+(0.5 \times 1.5-0.5 \times 1) \\F(x)=0.25+0.5+0.75-0.5 \\F(x)=1\end{array}F(x) from answer(a) isF(x)=\left\{\begin{array}{c}0 \quad \text { for } x<-2 \\-0.25 x+0.5 ; \text { for }-2 \leqslant x<1 \\0.5 x+0.25 ; \text { for } 1 \leqslant x<1.5 \\1 \quad \text { for } 1.5 \leqslant x\end{array}\right. ...