Solution:

a.
Suppose Ara goes first. He can begin by placing a $5$ in the center of the board. This is the only space shared by both diagonals, so when the game is over, the two diagonal sums are guaranteed to add up to $1+2+3+4+5+5=20$.

For her first move, Bea must play some number $n$ in one of the four corner spaces, where $n=1,2,3,$ or $4$. Ara can then respond by playing $5-n$ in the opposite corner, completing a diagonal. (We know this move is still available, because $5-n$ is either $1,2,3,$ or $4$, and cannot be equal to $n$.)

After these three moves, the sum of the completed diagonal is $5+n+(5-n)=10$. Since the two diagonal sums will add up to $20$ at the end of the game, the other diagonal sum will also be $10$. Thus Ara will win.

b.
Suppose Bea goes first. She can begin by placing a $2$ in the center of the board. By the same reasoning as in part (a), the two diagonal sums are now guaranteed to add up to $1+2+3+4+5+2=17$ at the end of the game.

For the diagonal sums to be equal, each diagonal sum would have to be $\frac{17}{2}$. But the diagonal sums are integers, so this cannot happen. Therefore, no matter what the remaining moves are, Bea will win.