Answer
a, $C=\frac{1}{2}$
b, $\frac{3}{8}$
c, $\frac{5}{48}$
Work Step by Step
a, We have: $$\int\int_{R^2}f(x,y)dA=\int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y)dxdy=1$$
But,
$$\int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y)dxdy = \int_{-\infty}^\infty\int_{-\infty}^\infty Cx(1+y)dxdy
\\=\int_0^1\int_0^2Cx(1+y)dydx
\\=\int_0^1Cxdx\int_0^2(1+y)dy
\\=\biggr[\frac{1}{2}Cx^2\biggr]_0^1\biggr[y+\frac{1}{2}y^2\biggr]_0^2
\\=\biggr[\frac{1}{2}C1^2\biggr]\biggr[2+\frac{1}{2}2^2\biggr]
\\=2C$$
Then $2C=1$ or $C=\frac{1}{2}$
b,$$P(X\leq1,Y\leq1)=\int_0^1\int_0^1\frac{1}2x(1+y)dxdy
\\=\int_0^1\frac{1}2xdx\int_0^1(1+y)dy
\\=\biggr[\frac{1}4x^2\biggr]_0^1\biggr[y+\frac{1}2y^2\biggr]_0^1
\\=\frac{1}41^2\biggr(1+\frac{1}21^2\biggr)
\\=\frac{3}8$$
c,$$P(X+Y\leq1)=\int_0^1\int_0^{1-y}\frac{1}2x(1+y)dxdy
\\=\int_0^1\biggr[\frac{1}4x^2(1+y)\biggr]_{x=0}^{x=1-y}dy
\\=\int_0^1\frac{1}4(1-y)^2(1+y)dy
\\=\int_0^1\frac{1}4(1-y^2)(1-y)dy
\\=\int_0^1\frac{1}4(y^3-y^2-y+1)dy
\\=\frac{1}4\biggr[\frac{1}4y^4-\frac{1}3y^3-\frac{1}2y^2+y\biggr]_0^1
\\=\frac{1}4\biggr(\frac{1}4-\frac{1}3-\frac{1}2+1\biggr)
\\=\frac{5}{48}$$
