Answer (1)$\dfrac{63}{256}$ (2)$\dfrac{99}{512}$

Solution to part 1 (Part 2 is similarly solved): Let $H$ represents heads and $T$ represent tails. There are $2^5$ possible ordered 5-tuples of $H$'s and $T$'s for 5 throws of a coin. The probability that $X=Y=k$ heads is equal to $P(k)^2$ where $P(k)$ denotes the probability of tossing $k$ heads in 5 throws. From this, we can do casework. For example, the probability of $0$ heads is ${5 \choose 0}\frac{1}{2^5}=\frac{1}{2^5}$. Thus, the probability that $X=Y=0$ is $\left(\frac{1}{2^5}\right)^2.$ It is well-known that the probability of throwing $k$ heads in 5 tosses of a fair coin is ${5 \choose k}*\frac{1}{2^5}$. Thus, we want to find the following sum: $\sum_{i=0}^5 \left({5 \choose k}*\frac{1}{2^5}\right)^2$. This is trivially computed by hand as $\frac{63}{256}$.