In attempting the following questions I am stuck and feel that both parts are the same...Anyone give hints on how I can proceed?
(a) Suppose that $X$ has a binomial distribution with parameters $(n, \theta)$ and that $Y$ is independent of $X$ and has a binomial distribution with parameters $(m, \theta)$, where $m$ and $n$ are known.
(i) [4 marks] Show that $T=X+Y$ is a sufficient statistic for $\theta$.
I think I can show this by either deriving that $T$ is a sufficient statistics of the likelihood for some $Z \in B(n+m,\theta)$ or show that the conditional distribution of n+m given $T(n+m)$ does not depend on $\theta$.
I will try the second method as I think that also answers (ii):
$$f(n+m\mid T(n+m,\theta))=\frac{f(n+m)}{f(T(n+m))}$$
$$f(n+m)=\prod_{}^{n+m}\binom{n+m}{k}\theta^k(1-\theta)^{n+m-k}$$
$$f(T(n+m))=\binom{n}{k}\theta^k(1-\theta)^{n-k}+\binom{m}{k}\theta^k(1-\theta)^{m-k}$$
But I am not sure if that simplifies.
(ii) [3 marks] Verify that the conditional distribution of $X$ and $Y$ given $T$ does not depend on $\theta$.
If the conditional distribution $(X,Y)\mid T$ is independent of $\theta$, then that already means $T$ is sufficient for $\theta$. So proving (ii) would prove (i).
For proving (i) separately, appeal to Factorisation theorem:
Distribution of $(X,Y)$ is
\begin{align} P(X=x,Y=y)&=P(X=x)P(Y=y) \\&=\begin{cases}\binom{n}{x}\binom{m}{y}\theta^{x+y}(1-\theta)^{n-x+m-y}&,\text{ if }x=0,1,\ldots,n;y=0,1,\ldots,m \\ 0&,\text{ otherwise}\end{cases} \\\\&=\begin{cases}\binom{n}{x}\binom{m}{y}\left(\frac{\theta}{1-\theta}\right)^{x+y}(1-\theta)^{n+m}&,\text{ if }x=0,1,\ldots,n;y=0,1,\ldots,m \\ 0&,\text{ otherwise}\end{cases} \\&=g(\theta,t)h(x,y) \end{align}
, where $g(\theta,t)=\left(\frac{\theta}{1-\theta}\right)^{x+y}(1-\theta)^{n+m}$ depends on $\theta$ and on $(x,y)$ through $t(x,y)=x+y$ and $h(x,y)$ is independent of $\theta$. This proves $T(X,Y)=X+Y$ is sufficient for $\theta$.
Now you know that $T=X+Y\sim\mathsf{Bin}(n+m,\theta)$. So for (ii), find this conditional probability for each $t=0,1,\ldots,n+m$ and show that it does not depend on $\theta$:
$$P(X=x,Y=y\mid T=t)=\begin{cases}\frac{P(X=x,Y=y)}{P(T=t)}&,\text{ if }t=x+y \\ 0&,\text{ otherwise }\end{cases}$$