Probability - Peculiar die 
A peculiar die has the following properties: on any roll the probability of rolling either a $4$, a $6$, or a $1$ is $1/2$, just as it is with an ordinary die. Moreover, the probability of rolling either a $1$, a $3$, or a $2$ is again $1/2$. However, the probability of rolling a $1$ is $5/16$, not $1/6$ as one would expect of an ordinary fair die.
From what you know about this peculiar die, compute the following:
a.    The probability of rolling either a $5$, a $3$, or a $2$;
b.    The probability of rolling a $5$.

My answer
Given $P(1) = \frac{5}{16}$, then
$$P(2) + P(3) + P(1) = P(2) + P(3) + \frac{5}{16} = \frac{1}{2} \iff P(2) + P(3) = \frac{3}{16}$$
Similarly, using $P(4) + P(6) + P(1) = \frac{1}{2}$, I find
$$P(6) + P(4) = \frac{3}{16}$$
b. $P(5) = 1 - \left(\frac{5}{16} + \frac{3}{16} + \frac{3}{16}\right) = \frac{5}{16}$
a. $P(5) + P(3) + P(2) = \frac{5}{16} + \frac{3}{16} = \frac{1}{2}$
Are my calculations correct?

I agree with your solution. I wrote a very similar solution below.
Let $E_j$ be the event that we roll a $j$. Then
$$
P(E_2 \cup E_3) = P(E_1 \cup E_2 \cup E_3) - P(E_1) = \frac12 - \frac{5}{16} = \frac{3}{16}.
$$
It follows that
\begin{align}
P(E_5) &= 1 - P(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_6) \\
&= 1 - P(E_1 \cup E_4 \cup E_6) - P(E_2 \cup E_3) \\
&= 1 - \frac12 - \frac{3}{16} = \frac{5}{16}.
\end{align}
Finally, we can put these pieces together to see that
$$
P(E_2 \cup E_3 \cup E_5) = P(E_2 \cup E_3) + P(E_5) = \frac{3}{16} + \frac{5}{16} = \frac12.
$$
