Step 1: Use the rule that probabilities add to 1.
The values are $P(0)=\tfrac{5}{16}$, $P(1)=\tfrac{5}{16}$, $P(2)=\tfrac{2k}{48}=\tfrac{k}{24}$, $P(3)=\tfrac14$. Their sum must be $1$.
Step 2: Solve for $k$.
$\tfrac{5}{16}+\tfrac{5}{16}+\tfrac{4}{16} = \tfrac{14}{16} = \tfrac78$, so $\tfrac{k}{24} = 1-\tfrac78 = \tfrac18$, giving $k = 3$.
Step 3: Write all probabilities over 16.
Now $P(2) = \tfrac{3\cdot 2}{48} = \tfrac18 = \tfrac{2}{16}$ and $P(3) = \tfrac{4}{16}$.
Step 4: Compute the mean.
\[ E(X) = 0\cdot\tfrac{5}{16} + 1\cdot\tfrac{5}{16} + 2\cdot\tfrac{2}{16} + 3\cdot\tfrac{4}{16} = \tfrac{5+4+12}{16} = \tfrac{21}{16} = 1.3125. \]
\[ \boxed{E(X) = 1.3125} \]