Apply the Remainder Theorem for the first condition.

According to the Remainder Theorem, if $f(x)$ is divided by $x-a$, the remainder is $f(a)$.
For division by $x+1$, we have $a = -1$. The remainder is given as 5.
So, $f(-1) = 5$.
Substitute $x = -1$ into $f(x) = 2x^3 + px^2 + qx + 1$:

$$2(-1)^3 + p(-1)^2 + q(-1) + 1 = 5$$ $$-2 + p - q + 1 = 5$$ $$p - q - 1 = 5$$ $$p - q = 6 \quad (*)$$

For division by $x-2$, we have $a = 2$. The remainder is given as 5.
So, $f(2) = 5$.
Substitute $x = 2$ into $f(x) = 2x^3 + px^2 + qx + 1$:

$$2(2)^3 + p(2)^2 + q(2) + 1 = 5$$ $$2(8) + 4p + 2q + 1 = 5$$ $$16 + 4p + 2q + 1 = 5$$ $$4p + 2q + 17 = 5$$ $$4p + 2q = 5 - 17$$ $$4p + 2q = -12$$

Divide by 2:

$$2p + q = -6 \quad (**)$$

We have two equations:

1. $p - q = 6$
2. $2p + q = -6$

Add equation (1) and equation (2) to eliminate $q$:

$$(p - q) + (2p + q) = 6 + (-6)$$ $$3p = 0$$ $$p = 0$$

Using equation (1):

$$p - q = 6$$ $$0 - q = 6$$ $$-q = 6$$ $$q = -6$$

The values are $p=0$ and $q=-6$.

The values of $p$ and $q$ are $\boxed{p=0, q=-6}$.