There are $4! = 24$ ways to arrange the ordering of the groups. For any winner/runner-up pairing, there are two ways it may occur. This means that at random, you have $\frac{2}{24} = \frac{1}{12}$ probability of getting it right.
Another way of looking at it is there are 4 choices for the winner, and consequently 3 choices for the runner-up (or vice-versa). Therefore there is a $\frac14 \cdot \frac13 = \frac{1}{12}$ chance you will pick both right randomly.
But, we must also independently consider the cases where you get only one answer right.
In this case, suppose you pick the winner right. There are 6 such ways to do that, but two of them, as we have noticed, also involve picking the runner-up correctly. As such, there are 4 ways to pick the winner right but not the runner-up, and vice versa. So this has a $\frac{4}{24} = \frac{1}{6}$ chance of happening.
The expected value of a single group is computed using the expected value formula:
$$\begin{align*}E[G_i] &= \underbrace{\frac{1}{12} \cdot 2}_{\textrm{picking both correctly}} + \underbrace{\frac{1}{6} \cdot 1}_{\textrm{picking only winner correctly}} + \underbrace{\frac{1}{6} \cdot 1}_{\textrm{picking only runner-up correctly}} \\
&= \frac16 + \frac16 + \frac16 \\
&= \frac12 \end{align*}$$
Since expectation is linear,
$$E\left[ \sum_{i=i}^8 G_i\right] = \sum_{i=1}^8 E[G_i] = \sum_{i=1}^8 \frac12 = 8\cdot \frac12 = 4.$$
Suppose we made the pool more interesting, and we gave 2 points for picking the winner, and 1 for the runner up. Then we can use this approach with only trivial modification:
$$\begin{align*}E[G_i] &= \underbrace{\frac{1}{12} \cdot 3}_{\textrm{picking both correctly}} + \underbrace{\frac{1}{6} \cdot 2}_{\textrm{picking only winner correctly}} + \underbrace{\frac{1}{6} \cdot 1}_{\textrm{picking only runner-up correctly}} \\
&= \frac14 + \frac13 +\frac16 \\
&= \frac34\\
E\left[\sum G_i\right] &= 8\cdot \frac34 \\
&= 6.
\end{align*}$$
