Assuming all points in the triangle [tex]T[/tex] are uniformly distributed, we have the joint density
[tex]f_{X,Y}(x,y)=\begin{cases}\frac1A&\text{for }(x,y)\in T\\0&\text{otherwise}\end{cases}[/tex]
where [tex]A=\frac{8^2}2=32[/tex] is the area of the triangle [tex]T[/tex].
(a)
[tex]P(X<5)=\displaystyle\iint_{T^*}f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx[/tex]
(where [tex]T^*[/tex] is the portion of [tex]T[/tex] for which [tex]x[/tex] is between 0 and 5)
[tex]P(X<5)=\displaystyle\frac1{32}\int_0^5\int_0^{8-x}\mathrm dy\,\mathrm dx[/tex]
[tex]P(X<5)=\displaystyle\frac1{32}\int_0^5(8-x)\,\mathrm dx[/tex]
[tex]P(X<5)=\dfrac1{32}\cdot\dfrac{55}2=\boxed{\dfrac{55}{64}}[/tex]
(b) Generalizing the previous result, we have
[tex]P(X\le x^*)=\displaystyle\iint_{T^*}f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx[/tex]
(this time with [tex]T^*[/tex] being the portion of [tex]T[/tex] where [tex]0\le x\le x^*[/tex] for some [tex]x^*[/tex] between 0 and 8)
[tex]P(X\le x^*)=\displaystyle\frac1{32}\int_0^{x^*}\int_0^{8-x}\mathrm dy\,\mathrm dx[/tex]
[tex]P(X\le x^*)=\displaystyle\frac1{32}\int_0^{x^*}(8-x)\,\mathrm dx[/tex]
[tex]P(X\le x^*)=\displaystyle\frac1{32}\left(8x^*-\frac{(x^*)^2}2\right)[/tex]
That is, the CDF of [tex]X[/tex] is
[tex]P(X\le x)=\begin{cases}\frac{8x-\frac{x^2}2}{32}&\text{for }0\le x\le8\\0&\text{otherwise}\end{cases}[/tex]
or
[tex]\boxed{P(X\le x)=\begin{cases}\frac{16x-x^2}{64}&\text{for }0\le x\le8\\0&\text{otherwise}\end{cases}}[/tex]
(c) Obtain the PDF by differentiating the CDF:
[tex]f_X(x)=\dfrac{\mathrm d}{\mathrm dx}P(X\le x)[/tex]
[tex]\boxed{f_X(x)=\begin{cases}\frac{8-x}{32}&\text{for }0<x<8\\0&\text{otherwise}\end{cases}}[/tex]
(d) Compute the expectation of [tex]X[/tex]:
[tex]E[X]=\displaystyle\int_0^8xf_X(x)\,\mathrm dx[/tex]
[tex]E[X]=\displaystyle\frac1{32}\int_0^8x(8-x)\,\mathrm dx=\boxed{\frac83}[/tex]
