To solve this problem, we need to find all strictly increasing functions $f: \mathbb{Z} \to \mathbb{Z}$ such that the condition given by:

$$\mho(f(a) - f(b)) \le \mho(a - b)$$

holds for all integers $a$ and $b$ with $a > b$.

### Step-by-step Solution:

1. Understand the Strictly Increasing Condition:
- Since $f$ is strictly increasing, for $a > b$, we have $f(a) > f(b)$.

2. **Analyzing $\mho$ Function**:
- The function $\mho(n)$ computes the sum of the exponents of prime factors of $n$ that are greater than $10^{100}$.
- For the inequality $\mho(f(a) - f(b)) \leq \mho(a - b)$, $f(a) - f(b)$ must have "less complex" prime factors (in the sense of being smaller or having smaller exponent multiplicities) compared to $a - b$.

3. Considering a Linear Function:
- A natural guess for a strictly increasing function from $\mathbb{Z}$ to $\mathbb{Z}$ is a linear function of the form $f(x) = Rx + c$, where $R$ and $c$ are integers.
- For linear functions, $f(a) - f(b) = R(a-b)$.

4. **Evaluate $\mho$ with Linear $f$**:
- Substitute into the inequality: $\mho(R(a-b)) \leq \mho(a-b)$.
- Given that $\mho(R(a-b))$ only considers primes greater than $10^{100}$, and if $R$ does not introduce any prime factor greater than $10^{100}$, then the inequality holds trivially.

5. Conclusion:
- Hence, any linear function $f(x) = Rx + c$ with integer $R$ that ensures $R$ has no prime factors greater than $10^{100}$ satisfies the condition.
- The general form of the solution is:
$$f(x) = Rx + c$$
where $R$ and $c$ are integers, and the prime factors of $R$ are all less than or equal to $10^{100}$.

Therefore, the strictly increasing functions satisfying the condition are expressed by:

$$\boxed{f(x) = Rx + c}$$
where $R$ is a positive integer whose prime factors do not exceed $10^{100}$, and $c$ is any integer.