Forming probability distributions of type $W=X+Y$. 
I am able to do part a) easily. My only problem here is finding the prob. distribution for $w=x+y$. If I could get that, I can solve c) also. Also, will it be the same case for $w=x-y$? Theres a similar question asking to find for $x-y$, but I'm guessing its the same case, so if I just know $x+y$, should be ok. So I just need answer for part (b).
 A: You can probably see that the possible values of $X+Y$ are $0,2,4,6,8,10$.
Call $X+Y$ by some name, like $W$. We want to find $\Pr(W=w)$ for $w=0,2,4,\dots,10$.
Unfortunately we have to do them one at a time.
What is $\Pr(W=0)$? This happens only if $X=0$ and $Y=0$. By independence, the probability of this is $(0.3)(0.4)=0.12$.
One down, quite a few to go.
What is the probability that $W=2$? This can happen in two ways, $X=0$, $Y=2$ or $X=2$, $Y=0$. So the probability is $(0.3)(0.2)+(0.4)(0.2)$.
Continue. Note that for $W=4$ there will be three different ways, for $6$ there will be also three  ways, but then things get easier. 
It will be easy to make a numerical slip or two. Note that $6$ numbers we get should add up to $1$. That gives a check on your calculations.
A: Only to expand on the previous answer, the probability distribution of the sum of two independent discrete random variables (like this) is actually the discrete convolution of the two probability mass functions. When writing the formula out:
$$\Pr(X+Y=w) = \sum_{k=-\infty}^{\infty} \Pr(X=w-k) \times \Pr(Y=k)$$
And for your other question about distribution for $X-Y$, the resultant mass function is the (discrete) cross-correlation:
$$\Pr(X-Y=w) = \sum_{k=-\infty}^{\infty} \Pr(X=w+k) \times \Pr(Y=k)$$
