= Solution
An <additive arithmetic function> satisfies $f(mn)=f(m)+f(n)$ for <coprime integers> $m,n$; a <completely additive arithmetic function> satisfies this for all positive $m,n$. In either case $f(1)=0$. The <fundamental theorem of arithmetic> gives $f(n)=\sum_{p^k\parallel n}f(p^k)$, where $p^k\parallel n$ means that the <prime valuation> of $n$ at $p$ is exactly $k$.
Under the <discrete uniform distribution> on $[N]$, the <expectation> of the <indicator function> of $p^k\parallel n$ is
$$
\frac1N\left(\left\lfloor\frac N{p^k}\right\rfloor-\left\lfloor\frac N{p^{k+1}}\right\rfloor\right)=\frac1{p^k}\left(1-\frac1p\right)+O(N^{-1}).
$$
The formula remains valid when $p^{k+1}>N$. Applying <linearity of expectation> to the finite <prime power> decomposition proves \b[the required mean formula]:
$$
\boxed{\mathbb E_N f=\sum_{p^k\le N}\frac{f(p^k)}{p^k}\left(1-\frac1p\right)+O\left(\frac1N\sum_{p^k\le N}|f(p^k)|\right).}
$$
For the last assertion, write the <excess prime-factor multiplicity> directly as a sum of nonnegative <indicator functions>:
$$
\Omega(n)-\omega(n)=\sum_p\sum_{k\ge2}\mathbf1_{p^k\mid n}.
$$
Here $\omega$ is the <prime omega function> and $\Omega$ is the <total number of prime factors>. The <expectation> is bounded independently of $N$, since
$$
\mathbb E_N(\Omega-\omega)\le\sum_p\sum_{k\ge2}\frac1{p^k}=\sum_p\frac1{p(p-1)}\le\sum_{m=2}^\infty\frac1{m(m-1)}=1.
$$
Thus <Markov inequality> gives \b[vanishing probability on every diverging scale]:
$$
\boxed{\mathbb P_N(\Omega-\omega\ge t(N))\le\frac1{t(N)}\longrightarrow0.}
$$
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