(a) The region Y₁ < 1/2 and Y₂ > 1/4 corresponds to the rectangle,
{(y₁, y₂) : 0 ≤ y₁ < 1/2 and 1/4 < y₂ ≤ 1}
Integrate the joint density over this region:
[tex]P\left(Y_1<\dfrac12,Y_2>\dfrac14\right) = \displaystyle\int_0^{\frac12}\int_{\frac14}^1 (y_1+y_2)\,\mathrm dy_2\,\mathrm dy_1 = \boxed{\dfrac{21}{64}}[/tex]
(b) The line Y₁ + Y₂ = 1 cuts the support in half into a triangular region,
{(y₁, y₂) : 0 ≤ y₁ < 1 and 0 < y₂ ≤ 1 - y₁}
Integrate to get the probability:
[tex]P(Y_1+Y_2\le1) = \displaystyle\int_0^1\int_0^{1-y_1}(y_1+y_2)\,\mathrm dy_2\,\mathrm dy_1 = \boxed{\dfrac13}[/tex]
Y₁ and Y₂ are not independent because
P(Y₁ = y₁, Y₂ = y₂) ≠ P(Y₁ = y₁) P(Y₂ = y₂)
To see this, compute the marginal densities of Y₁ and Y₂.
[tex]P(Y_1=y_1) = \displaystyle\int_0^1 f(y_1,y_2)\,\mathrm dy_2 = \begin{cases}\frac{2y_1+1}2&\text{if }0\le y_1\le1\\0&\text{otherwise}\end{cases}[/tex]
[tex]P(Y_2=y_2) = \displaystyle\int_0^1 f(y_1,y_2)\,\mathrm dy_1 = \begin{cases}\frac{2y_2+1}2&\text{if }0\le y_2\le1\\0&\text{otherwise}\end{cases}[/tex]
[tex]\implies P(Y_1=y_1)P(Y_2=y_2) = \begin{cases}\frac{(2y_1+1)(2y_2_1)}4&\text{if }0\le y_1\le1,0\ley_2\le1\\0&\text{otherwise}\end{cases}[/tex]
but this clearly does not match the joint density.
