Answer
$\color{blue}{p_X(k) = \begin{cases} 1/10, & k=2 \\ 2/10, & k=3 \\ 3/10, & k=4 \\ 4/10, & k=5 \end{cases}}$
$\color{blue}{p_V(k)= \begin{cases} 1/10, & k=3,4,8,9 \\ 2/10, & k=5,6,7 \end{cases}}$
Work Step by Step
The total number of possible outcomes when two balls are simultaneously drawn from an urn containing five (5) balls numbered $1, 2, 3, 4,$ and $5$ is $20$ as there are 5 possible outcomes for the first ball and 4 possible outcomes for the second ball (one ball has been previously drawn from the urn). In particular, note that it is not possible for both balls to have the same number. The outcomes are summarized in the tables below. The outcomes are equally likely and each has a probability of $1/20$ of occurring.
$\begin{array}{cc}
\underline{(B_1,B_2)} & \rm Ball\ 2\ (B_2) \\
\rm Ball\ 1\ (B_1)& \begin{array}{|c|c|c|c|c|} \hline
B_1\backslash B_2& 1 & 2 & 3 & 4 & 5 \\ \hline
1 & & (1,2) & (1,3) & (1,4) & (1,5) \\
2 & (2,1) & & (2,3) & (2,4) & (2,5) \\
3 & (3,1) & (3,2) & & (3,4) & (3,5) \\
4 & (4,1) & (4,2) & (4,3) & & (4,5) \\
5 & (5,1) & (5,2) & (5,3) & (5,4) & \\ \hline
\end{array}
\end{array}$
(a)
The table below lists the maximum number appearing when two balls are simultaneously drawn from the urn previously described. For example, if the outcome is $(B_1, B_2)=(2,3)$, that is, Ball 1 is 2 and Ball 2 is 3, then the maximum number appearing is $3$, and this is indicated in row 2 column 3 in the body of the table below. Note that the only possible maxima are 2, 3, 4, and 5.
$\begin{array}{cc}
\underline{\max\{B_1,B_2\}} & \rm Ball\ 2\ (B_2) \\
\rm Ball\ 1\ (B_1) & \begin{array}{|c|c|c|c|c|} \hline
B_1\backslash B_2 & 1 & 2 & 3 & 4 & 5 \\ \hline
1 & & 2 & 3 & 4 & 5 \\
2 & 2 & & 3 & 4 & 5 \\
3 & 3 & 3 & & 4 & 5 \\
4 & 4 & 4 & 4 & & 5 \\
5 & 5 & 5 & 5 & 5 & \\ \hline
\end{array}
\end{array}$
Since $X=\max\{B_1, B_2\}$, it can be easily seen from the table above that of the $20$ possible equally-likely outcomes, $X=2$ for 2 outcomes, $X=3$ for 4 outcomes, $X=4$ for 6 outcomes, and $X=5$ for 8 outcomes. Thus the pdf of $X$ is
$\begin{align*}
p_X(k) &= P(X=k),\ k=2,3,4,5 \\
&= P(\max\{B_1,B_2\} = k) \\
& \\
&= \begin{cases} 2/20, & k=2 \\ 4/20, & k=3 \\ 6/20, & k=4 \\ 8/20, & k=5 \end{cases} \\
& \\
\color{blue}{p_X(k)} &\color{blue}{= \begin{cases} 1/10, & k=2 \\ 2/10, & k=3 \\ 3/10, & k=4 \\ 4/10, & k=5 \end{cases}}
\end{align*}$
(b)
The table below lists the sums of the number appearing on the two balls from the urn previously described. For example, for the outcome $(B_1, B_2) = (2,4)$ (see row 2, column 4 in the body of the table below), the sum of the numbers appearing is $B_1+B_2= 2+4 = 6$. The only possible sums are 3, 4, 5, 6, 7, 8, and 9.
$\begin{array}{cc}
\rm B_1+B_2 & \rm Ball\ 2\ (B_2) \\
\rm Ball\ 1\ (B_1) & \begin{array}{|c|c|c|c|c|} \hline
B_1\backslash B_2 & 1 & 2 & 3 & 4 & 5 \\ \hline
1 & & 3 & 4 & 5 & 6 \\
2 & 3 & & 5 & 6 & 7 \\
3 & 4 & 5 & & 7 & 8 \\
4 & 5 & 6 & 7 & & 9 \\
5 & 6 & 7 & 8 & 9 & \\ \hline
\end{array}
\end{array}$
With $V=B_1+B_2$, it can be readily seen from the table that of the $20$ possible equally-likely outcomes, $V=3$ for 2 outcomes, $V=4$ for 2 outcomes, $V=5$ for 4 outcomes, $V=6$ for 4 outcomes, $V=7$ for 4 outcomes, $V=8$ for 2 outcomes, and $V=9$ for 2 outcomes. Thus, the pdf of $V$ is
$\begin{align*}
p_V(k) &= P(V=k),\ k=3,4,5,6,7,8,9 \\
&= P(B_1+B_2 = k) \\
& \\
&= \begin{cases} 2/20, & k=3 \\ 2/20, & k=4 \\ 4/20, & k=5 \\ 4/20, & k=6 \\ 4/20, & k=7 \\ 2/20, & k=8 \\ 2/20, & k=9\end{cases} \\
& \\
&= \begin{cases} 2/20, & k=3,4,8,9 \\ 4/20, & k=5,6,7 \end{cases} \\
& \\
\color{blue}{p_V(k)} &\color{blue}{= \begin{cases} 1/10, & k=3,4,8,9 \\ 2/10, & k=5,6,7 \end{cases}}
\end{align*}$
