Joint probability distribution $Y_1$ and $Y_2$ are jointly distributed with density $f(y_1,y_2)=4y_2^2 \qquad 0 \leq y_1 \leq y_2 \leq 1$
Determine the following:
$P(\text{max} \{Y_1,Y_2\} <1/2) = \int_{y_2=0}^{1/2}\int_{y_1=0}^{y_2}4y_2^2dy_1dy_2  \approx 0.0625$ 
$P(Y_1+Y_2 < 1/2) = \int_{y_1=0}^{1/4}\int^{1/2-y_1}_{y_2=y_1}4y_2^2dy_2dy_1 \approx 0.0182292$ 
$P(Y_1Y_2 <1/2) = \int_{y_2=0}^{\sqrt{1/2}}\int_{y_1=0}^{y_2}4y_2^2dy_1dy_2 + \int_{y_2=\sqrt{1/2}}^{1}\int_{y_1=0}^{1/(2y_2)}4y_2^2dy_1dy_2 = 1/4 + 1/2 = 0.75$ 
$P(Y_1/Y_2<1/2) = \int_{y_2=0}^{1}\int_{y_1=(1/2)y_2}^{y_2}4y_2^2dy_1dy_2 = 0.5$ 
$P(Y_2-Y_1 < 1/2) = \int_{y_2=0}^{1/2}\int_{y_1=0}^{y_2}4y^2_2dy_1dy_2 + \int_{y_2=1/2}^{1}\int_{y_1=y_2-(1/2)}^{y_2}4y^2_2dy_1dy_2 \approx 0.583333+0.0625 = 0.645833$ 
$P(\text{min} \{Y_1,Y_2\}<1/2) = \int_{y_2=1/2}^{1}\int_{dy_1=0}^{y_2}4y^2_2dy_1dy_2 \approx 0.9375 $ 

We do the first problem, since that is the one you attacked in some detail.   Subscripts are a pain to type. Also, the geometry of the $x$-$y$ plane is familiar. So we use $x$ instead of $y_1$ and $y$ instead of $y_2$, though reversing the choices would be better.
And we call the random variables $X$ and $Y$.
Our density function is then $4y^2$, and lives on $0\le x\le y\le 1$. This region is a triangle with corners $(0,0))$, $(1,1)$, and $(0,1)$. Draw the triangle. The boundary line $y=x$ will be important. 
We want the probability that $X\lt 1/2$ and $Y\lt 1/2$. This is the integral of the joint density function over the part of our triangle that is inside the square with corners $(0,0)$, $(1/2,0)$, $(1/2,1/2)$ and $(0,1/2)$. 
Draw the region over which we need to integrate. It is a triangle with corners $(0,0)$, $(1/2,1/2)$, and $(0,1/2)$.
Now we are finished. It is a matter of taste whether we integrate first with respect to $x$ or with respect to $y$. Let's do $x$ first. We want
$$\int_{y=0}^{1/2}\left(\int_{x=0}^y 4y^2\,dx                  \right)\,dy.$$
The others are quite similar, mostly a little harder. In all cases the only real work is to carefully identify the region over which we are integrating. That is why we deliberately spent so much time focusing on the geometry. Sometimes it is convenient to find first the probability of the complement.
For example, for $X+Y\lt 1/2$ we will draw our basic triangle again, and draw the line $x+y=1/2$. In this case it will be convenient to integrate first with respect to $y$.
If one of the problems causes difficulty, please leave a comment.
