Find the number of all sequences $\{ a_{n}\}$ in $\{-5,-4,-3,...,0,1,...,100\}$ such that $ |a_{n}| < |a_{n+1}|$. 
Find the number of all sequences $\{ a_{n}\}$ in $\{-5,-4,-3,...,0,1,...,100\}$ such that $ |a_{n}| < |a_{n+1}|$.

I think we can find a unique sequence for every non empty subset of $\{0,1,\ldots ,100\}$ so we have at least $2^{101}-1$ .

Let $\{a_n\}$ be a sequence [possibly empty]. By the reasoning of the OP, the sequence $\{a_n\}$ is completely determined by 1. and 2 together below:

*

*for each of the $5$ integers $i \in \{1,2,3,4,5\}$, whether $i,-i$, or neither, is in $\{a_n\}$  [so $3$ choices, and those are precisely the $3$ choices for each of those $5$ integers $i$],


*for each of the remaining $96$ integers $i \in \{0,6,7,8,\ldots, 100\}$, whether or not $i$ is in $\{a_n\}$  [so $2$ choices for each of those $96$ integers $i$].
So this yields precisely $3^5 \times 2^{96}$ choices for $\{a_n\}$, including the empty sequence. There is however, exactly $1$ empty sequence, the number of **nonempty $\{a_n\}$ is then $(3^5 \times 2^{96})-1$.
What if the condition were changed to $|a_n| \le |a_{n+1}|$, instead of $|a_n|<|a_{n+1}|$. Then the number of such sequences becomes $4^52^{96}-1$. Can you see why.
