$X_1,X_2,...,X_n$ follows iid $Bernoulli(p)$, $0<p<1$
Let $$ \tau(p)={(p+ qe^3)}^2 $$ is a function of parameter $p$ where $q=1-p$.
Q.1 $\:$ Write an unbiased estimator $T$ for $\tau(p)$
Q.2 $\:$ If $U=\sum_{i=1}^{n}X_i$ is the complete sufficient statistic, then calculate $E_p[T|U=u]$ using Lehmann-Schefee theorems
Q.3 $\:$ Find the UMVUE for $\tau(p)$.
($Hint:$ Use the concept of mgf of the Xs)
$\underline{\Large Solution}$:
I have chosen $T=e^{3(1-X_1)} \: e^{3(1-X_2)}$ which is unbiased for $\tau(p)$.
I have calculated $E_p[T|U=u]$ which is $$E_p[T|U=u]= e^6\: \frac{(n-u)(n-u-1)}{n(n-1)} + 2 e^3 \: \frac{u(n-u)}{n(n-1)} + \frac{u(u-1)}{n(n-1)} $$
Now my questions are :
i.) Are my founding results correct?
ii.) Is it possible to simplify the resulting quantity of $E_p[T|U=u]$?
iii.) What is the answer to Q.3?
iv.) Please show me the Expectation of at least one of the three parts of $E_p[T|U=u]$
v.) Should I choose $T=[X_1 + (1-X_1)e^3] \: [X_2 + (1-X_2)e^3]$ ?
Your procedure is correct and you calculations look fine to me.
I did check only the first addend of your conditional expectation:
$$\mathbb{P}[T=e^6|U=u]=\frac{\mathbb{P}[T=e^6]\mathbb{P}[\sum_{i=1}^{n}X_i=u|X_1=0,X_2=0]}{\mathbb{P}[\sum_{i=1}^{n}X_i=u]}=$$
$$=\frac{q^2\binom{n-2}{u}p^u q^{n-2-u}}{\binom{n}{u}p^uq^{n-u}}=\dots=\frac{(n-u)(n-u-1)}{n(n-1)}=(1-\overline{X}_n)(1-\frac{n}{n-1}\overline{X}_n)$$
Which is exactly your result and thus I have no doubts that the rest is correct too.
Finally calculate
$$\mathbb{E}[T|U]=1\cdot\mathbb{P}[T=1|U=u]+e^3\cdot\mathbb{P}[T=e^3|U=u]+e^6\cdot\mathbb{P}[T=e^6|U=u]$$
Q3 Using Lehmann Scheffé and Rao Blackwell results together, $\mathbb{E}[T|U]$ is UMVUE for $\tau(p)$