when solving (b)
Is variance $$V(\frac{1}{2}(x_1+x_2)) = \frac{1}{4}V(x_1+x_2)= \frac{1}{4}(v(x_1)+v(x_2))= \frac{1}{2}\sigma^2$$ or should I divide variance by the sample size so that $$V(\frac{1}{2}(x_1+x_2))= \frac{1}{4}(v(x_1)+v(x_2)= \frac{1}{4}(\frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2})$$
For (a), $\mathbb E\left[\hat\mu\right] = \mathbb E\left[\tilde\mu\right] = \mu$, so both estimators are unbiased.
For (b), we compute
$$\begin{align*} \operatorname{Var}\left(\hat\mu\right) &= \operatorname{Var}\left(\frac12\bar X_1+\frac12\bar X_2\right)\\ &=\frac14\operatorname{Var}\left(\bar X_1\right) + \frac14\operatorname{Var}\left(\bar X_2\right)\\ &= \frac1{4n_1}\sigma^2 + \frac1{4n_2}\sigma^2\\ &= \left(\frac{n_1+n_2}{4(n_1n_2)}\right)\sigma^2, \end{align*}$$ and $$\begin{align*} \operatorname{Var}\left(\tilde\mu\right) &= \operatorname{Var}\left(\frac{n_1\bar X_1 + n_2\bar X_2}{n_1+n_2}\right)\\ &= \left(\frac{n_1}{n_1+n_2}\right)^2\operatorname{Var}\left(\bar X_1\right) + \left(\frac{n_2}{n_1+n_2}\right)^2\operatorname{Var}\left(\bar X_2\right) \\ &= \frac{n_1}{(n_1+n_2)^2}\sigma^2 \frac{n_2}{(n_1+n_2)^2}\sigma^2\\ &= \left(\frac{n_1+n_2}{(n_1+n_2)^2}\right)\sigma^2 \end{align*}$$
So $\hat\mu$ has lower variance when $4n_1n_2>(n_1+n_2)^2$, otherwise $\tilde\mu$ has lower variance.