I am sorry if the terminology is a little bit off, and anyone that knows the correct terminology please correct me.
Let us assume we have 3 independent measures, X1, X2, and X3 from the same population with expectation μ and standard deviation σ. Which of the two estimators are best?
Apparantly they are both, expectation right. However, I don't understand how they conclude that. They are just pointing me to a formula, saying this is easy.
I understand how to see which of them have the higher variance though.
Another example:
A random person have height with unknown expectation μ and standard deviance σ. Let's choose two different, independent persons, and measure their heights that we will call X1, and X2. Calculate their expectation value and variance to the following estimators for μ. What estimator is the best one? The third one is not expectation right.
The formula $$ E(a_1X_1 + a_2X_2 + \cdots + a_nX_n+b)=a_1E(X_1)+a_2E(X_2)+\cdots+a_n(X_n)+b\tag{*} $$ is all you need to determine if an estimator is unbiased. For your first example, we expand out $\hat\mu_1$ as follows: $$ \textstyle\hat\mu_1=\frac13(X_1+X_2+X_3)=\frac13X_1+\frac13X_2+\frac13X_3 $$ and then apply ($*$): $$ \textstyle E(\hat\mu_1)=E(\frac13X_1+\frac13X_2+\frac13X_3)\stackrel{(*)}=\frac13E(X_1)+\frac13E(X_2)+\frac13E(X_3)=\frac13\mu+\frac13\mu+\frac13\mu=\mu. $$ Since $\hat\mu_1$ has expectation equal to $\mu$, this means that $\hat\mu_1$ is an unbiased estimator for $\mu$.
Similarly you can calculate $E(\hat\mu_2)$ by first expanding $$ \textstyle\hat\mu_2=\frac16(X_1+2X_2+3X_3)=\frac16X_1+\frac26X_2+\frac36X_3. $$ Using ($*$) again this gives $$ \textstyle E(\hat\mu_2)\stackrel{(*)}=\frac16E(X_1)+\frac26E(X_2)+\frac36E(X_3)= \mu, $$ so $\mu_2$ is an unbiased estimator for $\mu$.
You should be able to handle the second set of problems in the same way!