b) $a_4+a_6 \geq 2a_5$ So $a_4<a_5,a_6<a_5$ is impossible

a) Let $a$ be sufficiently large positive real number greater than $1$ that $a^9(a^2-1)>1$ and let $b=\frac{1}{a}<1$. Then, consider given numbers $a,\underbrace{b,b,...,b}_{n}$ where $n$ is a positive integer that $\frac{a^5-a^4}{b^4-b^5}<n<\frac{a^6-a^5}{b^5-b^6}$. Note that such $n$ exists since our choice of $a$ guarantee that the interval length is greater than $1$. We've $a_i=a^i+nb^i$ for all positive integer $i$. Keep in mind that $a>1>b>0$, we've $a_i>a_{i+1}\iff n>\frac{a^{i+1}-a^i}{b^i-b^{i+1}}$ and $a_i<a_{i+1}\iff n<\frac{a^{i+1}-a^i}{b^i-b^{i+1}}$. Also, we've $\frac{a^{i+1}-a^i}{b^i-b^{i+1}} <\frac{a^{i+2}-a^{i+1}}{b^{i+1}-b^{i+2}}$ for all positive integer $i$. Not hard to see that our choice of $n$ guarantee both $a_i>a_{i+1}$ for all $i\in [4]$ and $a_i<a_{i+1}$ for all $i\in \mathbb{Z}_{\geq 5}$.

b) We'll prove that this case is impossible. We know that in this case max $\{a_{n}\} = a_{5} $. Suppose that $a_{4} = b_{1}^{4} + b_{2}^{4} + \cdots + b_{k}^{4}$ and that $ a_{6} = b_{1}^{6} + b_{2}^{6} + \cdots + b_{k}^{6} $ , so $$ a_{4} + a_{6} = b_{1}^{5} \left( b_{1} + \frac{1}{b_{1}} \right) + b_{2}^{5} \left( b_{2} + \frac{1}{b_{2}} \right) + \cdots + b_{k}^{5} \left( b_{k} + \frac{1}{b_{k}} \right) \ge 2a_{5}$$ Thus, the initial conditions $a_{4} < a_{5}$ ; $a_{6} < a_{5}$ are impossible.