Encoding an element in $r=r'||h(r')$ way We consider a finite field $\mathbb{F}_p$, where $p$ is a large prime e.g. 256-bit.

We have $b$ a fixed element of the field. We encode it as $b'=b||h(b)$, where $h(.)$ is a cryptographic hash function. Assume the output of $h(.)$ is of size 160-bit.
We encode that way to distinguish the element $b'$ from a random element $r$ of the field. 

So to check if the element has the above structure we do: 

(1) Parse the value: $r=r'||r''$, where size of $r''$ is 160-bit. 

(2) check: $h(r')\stackrel{?}=r''$



Question1: How can we show/prove that the random element only with a negligible probability can have such structure?

Question2: Can we reduce the $h(.)$ output size to 80-bit (which is not the standard hash function output size) and set $p$ as smaller prime number (e.g. 128-bit )and prove a random value can have the above structure with only negligible probability?

Citing a paper that use/prove such statement would suffice.
Question1: How can we show/prove that the random element only with a negligible probability can have such structure?


If we assume the random oracle model and have $h: \{0,1\}^* \to \{0,1\}^n$ then we can state that $h(.)$ is equivalent to randomly sampling from $\{0,1\}^n$. Thus for a random element $r = r'||r''$ the probability that $r$ has the structure $r'' = h(r')$ is $\frac{1}{2^n}$ since all $2^n$ possible outputs of $h(r')$ are equally likely. Thus for $n=160$ we have a probability of $\frac{1}{2^{160}}$ which is negligible.


  Question2: Can we reduce the $h(.)$ output size to 80-bit (which is not the standard hash function output size) and set $p$ as smaller prime number (e.g. 128-bit )and prove a random value can have the above structure with only negligible probability?


Sure, the same logic as above can be applied here as well, now with the case $n=80$ and a corresponding probability of $\frac{1}{2^{80}}$ which is still negligible. 

Note that the size of $r'$ is not relevant to the probability of a random element having the structure $r'||h(r')$, the probability solely depends on the size of $h(r')$.