WLOG $a_5$ is the greatest or the least in $(a_k)$. By C-S $100=a_5^2+\sum\limits_{k=1}^4a_k^2\ge a_5^2+\frac{1}{4}\left(\sum\limits_{k=1}^4a_k\right)^2=a_5^2+\frac{1}{4}(20-a_5)^2$, i.e. $(8-a_5)a_5\ge 0$, which says $\max(\max(a_k))=8$ for $(3,3,3,3,8)$ and $\min(\max(a_k))=5$ for $(5,5,5,5,0)$.

Notice that $$\sum_{k = 1}^{5}(a_k-3)^2 = 25 \implies a_k \leq 8$$and since $(3,3,3,3,8)$ is a solution, the maximum of the maximum is $8$. Suppose for the sake of contradiction that $a_i < 5$ for all $i=1,2,3,4,5$. Then summing the inequailites $5a_i \geq a_i^2$ for all $i=1,2,3,4,5$ gives that $(5-a_i)a_i=0$ for all $i$ which is clearly a contradiction. So the maximum is at least $5$ and $(0,5,5,5,5)$ is an equailty case.

In order to find the maximum, WLOG $a_5$ is the largest of $(a_1,a_2,a_3,a_4,a_5)$ and then obtain $\displaystyle{\sum_{k=1}^4 a_k = 20 - a_5}$ and also $\displaystyle{\sum_{k=1}^4 a_k^2 = 100 -a_5^2}$. By QM-AM we obtain \[4(100-a_5^2) \geq (20-a_5)^2 \Longrightarrow 0 \geq 5a_5^2 - 40a_5 \Longrightarrow 0 \geq a_5(a_5-8) \Longrightarrow 0 \leq a_5 \leq 8\]Now we can just construct the pair $(3,3,3,3,8)$ and we have proven that the maximum value is $\boxed{8}$. Now we'll define the minimum value to be $\alpha$ and we'll define $d_i$ for $1 \leq i \leq 5$ such that $a = \alpha - d_1$, $b = \alpha - d_2$ and so on. Notice we now have $\displaystyle{5\alpha - 20 = \sum_{k=1}^5 d_i}$ and $\displaystyle{5\alpha^2 - 40\alpha + 100 = \sum_{k=1}^5 d_i^2}$. However we know that $\displaystyle{\sum_{k=1}^5 d_i^2} \leq \displaystyle{(\sum_{k=1}^5 d_i)^2}$ and thus we have \[(5\alpha - 20)^2 \leq 5\alpha^2 - 40\alpha + 100 \Longrightarrow \alpha^2 - 8\alpha + 15 \geq 0 \Longrightarrow (\alpha - 3)(\alpha - 5) \geq 0\]Which implies either $\alpha \leq 3$ (which is clearly impossible), or $\alpha \geq 5$, which we can use to obtain the construction for the minimum value, namely $(5,5,5,5,0)$. Thus we obtain that the minimum value is $\boxed{5}$.