Now that there is another full answer, and the question is already fairly old, here is my approach. My favorite tool for logical problems is to use (equational) logic.
For knight/knave puzzles, we have the following axiom:
$
\newcommand{\says}[2]{#1\text{ says }\unicode{x201C}#2\unicode{x201D}}
\newcommand{\cansay}[2]{#1\text{ can say }\unicode{x201C}#2\unicode{x201D}}
$\begin{align}
\tag{0} & \says{x}{P} \;\Rightarrow\; (T(x) \equiv P) \\
\end{align}
Here $\;T(x)\;$ stands for "$\;x\;$ speaks the truth" or "$\;x\;$ is a knight", and similarly $\;\lnot T(x)\;$ stands for "$\;x\;$ lies" or "$\;x\;$ is a knave".
I will be spelling out the logical part quite a bit; hope that isn't a problem.
Note that $\;\equiv\;$ is associative, and we will use that below to make logical formulas simpler to read: so below, $\;P \equiv Q \equiv R\;$ stands for either $\;(P \equiv Q) \equiv R\;$ or $\;P \equiv (Q \equiv R)\;$, which are equivalent. Do not confuse this with $\;(P \equiv Q) \;\land\; (Q \equiv R)\;$.
(a) Suppose you come across two of the natives. You ask this question "whether the other one is a knight?" from each of them. Will you get the same answer in each case?
So calling them $\;x\;$ and $\;y\;$, and calling their respective answers $\;r\;$ and $\;s\;$, we are given that $\;\says{x}{T(y) \equiv r}\;$ and $\;\says{y}{T(x) \equiv s}\;$. What can we say about $\;r\;$ and $\;s\;$? Let's simply calculate:
\begin{align}
& \says{x}{T(y) \equiv r} \;\land\;\says{y}{T(x) \equiv s} \\
\Rightarrow & \qquad \text{"by (0), twice -- the only thing we can do"} \\
& (T(x) \equiv T(y) \equiv r) \;\land\; (T(y) \equiv T(x) \equiv s) \\
\equiv & \qquad \text{"logic: $\;\equiv\;$ is symmetrical"} \\
& (T(x) \equiv T(y) \equiv r) \;\land\; (T(x) \equiv T(y) \equiv s) \\
\Rightarrow & \qquad \text{"logic: $\;\equiv\;$ is transitive"} \\
& r \equiv s \\
\end{align}
So yes, the answers will always be the same.
(b) There are three natives A, B and C. Suppose A says "B and C are the same type". What can be inferred about the number of knights?
So we know $\;\says{A}{T(B) \equiv T(C)}\;$, in other words
\begin{align}
& \says{A}{T(B) \equiv T(C)} \\
\Rightarrow & \qquad \text{"by (0) -- the only thing we can do"} \\
& T(A) \equiv T(B) \equiv T(C) \\
(*) \;\;\; \equiv & \qquad \text{"logic: property of any 'chain' of $\;\equiv\;$"} \\
& \text{an even number of }T(A), T(B), T(C)\text{ is false} \\
\end{align}
In other words, there is an even number of knaves among $\;A,B,C\;$, and or equivalently an odd number of knights.
The property of $\;\equiv\;$ used in step $(*)$ can be shown in this case using a truth table, and in general by an induction argument.
(c) You would like to determine whether an odd number of A, B and C is a knight. You may ask one yes/no question to any one of them. What is the question you should ask?
Ah, that's a coincidence! We just discovered in (b) that "there is an odd number of knights among $\;A,B,C\;$" can be formalized as $\;T(A) \equiv T(B) \equiv T(C)\;$.
Therefore we want to find out a question $\;Q\;$ such that the reply is always yes (true) if there is an odd number, and no (false) if there is an even number. In other words, we want to find $\;Q\;$ such that (for any $\;x,r\;$)
$$
\says{x}{Q \equiv r} \;\Rightarrow\; T(A) \equiv T(B) \equiv T(C) \equiv r
$$
However, since there is nothing to indicate how $\;A,B,C\;$ can affect our solution, let's be bold and try to find such a $\;Q\;$ to determine the truth of any statement $\;P\;$:
$$
\says{x}{Q \equiv r} \;\Rightarrow\; P \equiv r
$$
Let's try to transform that last expression into one of the form $\;Q \equiv \ldots\;$:
\begin{align}
& \says{x}{Q \equiv r} \;\Rightarrow\; P \equiv r \\
\Leftarrow & \qquad \text{"by (0), using transitivity of $\;\Rightarrow\;$} \\
& \qquad \phantom{\text{"}}\text{-- the only thing we know about $\;\says{\cdot}{\cdot}\;$"} \\
& (T(x) \equiv Q \equiv r) \;\Rightarrow\; P \equiv r \\
\Leftarrow & \qquad \text{"weaken -- the simplest way I see to make progress"} \\
& T(x) \equiv Q \equiv r \equiv P \equiv r \\
\equiv & \qquad \text{"logic: $\;\equiv\;$ is symmetric, twice -- to achieve $\;Q \equiv \ldots\;$"} \\
& Q \equiv T(x) \equiv P \equiv r \equiv r \\
\equiv & \qquad \text{"logic: $\;\phi \equiv \phi \;\equiv\; \text{true}\;$; simplify"} \\
& Q \equiv T(x) \equiv P \\
\equiv & \qquad \text{"introduce abbreviation (1), see below"} \\
& Q \equiv \cansay{x}{P} \\
\end{align}
In the last step I introduced the abbreviation
$$
\tag{1} \cansay{x}{P} \;\equiv\; T(x) \equiv P
$$
that I could have used from the start, but it only becomes necessary here.
So to know anything $\;P\;$ in this world, including whether $\;A,B,C\;$ have an odd number of knights, you just ask, "Can you say $\;P\;$?", and "Yes" means that $\;P\;$ is true, "No" means it is not.
Now, can you say that's easy?
(d) There are two natives, A and B. Now A says, "B is a knight is the same as I am a knave". What can you determine about A and B?
The difficult thing here it to get the correct formalization, which is $\;\says{A}{T(B) \equiv \lnot T(A)}\;$. Now the answer is an easy calculation:
\begin{align}
& \says{A}{T(B) \equiv \lnot T(A)} \\
\Rightarrow & \qquad \text{"by (0) -- the only thing we can do"} \\
& T(A) \equiv T(B) \equiv \lnot T(A) \\
\equiv & \qquad \text{"logic: $\;\equiv\;$ is symmetrical"} \\
& T(A) \equiv \lnot T(A) \equiv T(B) \\
\equiv & \qquad \text{"logic: contradiction"} \\
& \text{false} \equiv T(B) \\
\equiv & \qquad \text{"logic: simplify"} \\
& \lnot T(B) \\
\end{align}
So the only thing we can say is that $\;B\;$ is a knave: we know nothing about $\;A\;$.
