In my first answer, I assumed all values are unknown to us. Now let's consider the scenario where we know the value of a secretary before making an offer.
Again, consider the 2-secretary scenario again. We do know $v_1$ before making the first offer, but not $v_2$. For the second secretary we again thus just offer our remaining budget.
The expected reward for the second secretary (not knowing their value) offered budged $b$ is $b^2/2+(1-b)b$ (see first answer).
Assume the first secretary has value $v_1$ and is offered budget $b$. W.l.o.g. we can assume $b\leq v_1$ here, since the optimal solution surely does not offer more than necessary for secretary 1 to surely accept.
We again consider two cases: Either secretary 1 accepts and offer of $b$, or secretary 1 declines.
The expected workload of secretary 1 is simply $v_1$ in the first scenario and $0$ in the second. The probabilities are $b/v_1$ and $1-b/v_1$, respectively.
For the second secretary, the expected value depends on whether the first secretary acceoted. If the first accepted, we have $(1-b)$ budget left and thus have an expected workload of $(1-b)^2/2+b(1-b)$. If the first secretary declined, the second surely accepts the offer of $1$ and has expected workload $1/2$.
So in total we get an expected workload of
$$ (b/v_1)(v_1+(1-b)^2/2+b(1-b))+(1-b/v_1)(0+1/2) $$
We can simplify this to $f(b)=-b^3/(2 v_1) + b + 1/2$, i.e., a simple polynomial in $b$. We want to find the maximum under the constraint $b\leq v_1$ as this is not explicitly handled in the term.
Taking the derivative, we find that $f(b)$ has two extrema at $b=\pm \sqrt{\frac{2}{3}v_1}$. The one at negative $b$ is a minimum, the one at positive $b$ a maximum. Between those, $f(b)$ is monotonically increasing. Thus, we simply have that if $\sqrt{\frac{2}{3}v_1}$ is within the interval $[0,v_1]$ we offer that value, otherwise we just offer $v_1$.
So the optimal strategy is:
$$\mathit{offer}(v_1)=\begin{cases}\sqrt{\frac{2}{3}v_1} & \text{ if } \sqrt{\frac{2}{3}v_1}\leq v_1 \\
v_1 & \text{ else }
\end{cases}$$
The condition is satisfied iff $v\geq 2/3$, so equivalently (and more readable):
$$\mathit{offer}(v_1)=\begin{cases}
v_1 & \text{ if }v_1 < 2/3 \\
\sqrt{\frac{2}{3}v_1} & \text{ if } v_1 \geq 2/3
\end{cases}$$
For the expected workloads in each case, we compute $f(v_1)$ and $f(\sqrt{\frac{2}{3}v_1})$:
$$
f(v_1)=-v_1^2/2 + v_1 + 1/2 \\
f(\sqrt{\frac{2}{3}v_1})=\sqrt{\frac{8}{27}v_1}+1/2
$$
In the first case, $v_1$ is uniformly distributed over $[0,2/3)$, in the second over $[2/3,1]$. Taking the integrals over the respective intervals we get the final expected workload:
$$ \int_0^{2/3} -v_1^2/2 + v_1 + 1/2 dv_1 + \int_{2/3}^1 \sqrt{\frac{8}{27}v_1}+1/2 d_v1 = 0.506...+0.322...=0.838... $$
For generalization to more secretaries, we can go back to my first answer. If we do not know $v_1$, we can still compute an expected value for a budget $B$. We can do the same generalization for the case where we know the first value by adapting the computation above.
In the 3-secretary scenario we can then simply compute the expected value when offering $b$ to the first secretary by considering the expected value for the remaining two secretaries when the starting budget is $1-b$, and then optimize for $b$ and we're done. When we observe $v_2$ we can then adapt the strategy to optimize for that.
Now for completeness there's still the case where we know ALL values beforehand... :)
