= Solution
The <Lehmann–Scheffé theorem> states that an unbiased estimator which is a function of a complete sufficient statistic is the unique minimum-variance unbiased estimator; equivalently conditioning any unbiased estimator on such a statistic produces it.
Use the unbiased estimator $I=\mathbf1_{X_1>x}$. Given $T_n=t$, the positive sample vector is uniform on the simplex $\sum_iX_i=t$: the joint exponential density is constant there, independently of $\lambda$. For $t<x$ the conditional probability is zero. For $t\geq x$, translating the first coordinate by $x$ changes the simplex size from $t$ to $t-x$; its $(n-1)$-dimensional volume ratio is $(1-x/t)^{n-1}$. Thus
$$
\mathbb E[I\mid T_n]=\widetilde\phi_n,
$$
including $n=1$, when this is the indicator $\mathbf1_{T_1\geq x}$ up to a null boundary. <Conditional expectation> preserves unbiasedness, and completeness from part (i) proves \b[the displayed estimator is the unique minimum-variance unbiased estimator.]
It does not attain the <Cramér-Rao bound>. Equality in that bound requires the centred unbiased estimator to be proportional to the <score function>, hence affine in $T_n$ at fixed $\lambda$. This estimator is zero on a positive-probability interval and positive elsewhere, so is not affine; the $n=1$ step function is not affine either.
For the last assertion, the remaining observations are independent of the first $k$. Conditional on $T_k$, the internal proportions of those first $k$ are independent of the remaining sum, so
$$
\mathbb E[I\mid T_k,T_n]=\widetilde\phi_k.
$$
Now the tower property for the nested information $\sigma(T_n)\subseteq\sigma(T_k,T_n)$ gives
$$
\boxed{\mathbb E[\widetilde\phi_k\mid T_n,\lambda]=\mathbb E[I\mid T_n,\lambda]=\widetilde\phi_n.}
$$
Conditioning first on both sums is essential; the sigma-fields generated by $T_k$ and $T_n$ alone are not nested.
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