For the following statements, I need to either prove it is true, or find a counterexample to demonstrate it is false.
Here is what I have done for each one:
(a) If $f: \mathbb R \to \mathbb R$ is an even function, then the function $g: \mathbb R \to \mathbb R$ defined by $g(x) = -f(x)$ is also an even function.
True.
$$\begin{align} g(x) &= -f(x) \\ g(-x) &= -f(-x) \\ &= -f(x) \\ &= g(x) \text{ which is even.} \end{align}$$
(b) If $f: \mathbb R \to \mathbb R$ is an odd function, then $f$ is injective.
False.
Let $f(x) = \sin(x).$
$$\begin{align} f(-x) &= \sin(-x) \\ &= -\sin(x) \\ &= -f(x) \end{align}$$
$\sin(x)$ is odd but not injective.
(c) For all real numbers $x \in [0, 1],$ $\sin^{-1} x + \cos^{-1} x = \pi/2.$
True.
Let $\sin^{-1}(x) = \theta.$
$$\begin{align} x &= \sin\theta \\ &= \cos(\pi/2 - \theta) \\ \cos^{-1} x &= \pi/2 - \theta \\ \cos^{-1} x &= \pi/2 - \sin^{-1} x \\ \cos^{-1} x + \sin^{-1} x = \pi/2 \end{align}$$ Are these correct?
Yes your answers are all correct.
A couple of things though:
For the last line of (a), I would write "$ =g(x).$ Therefore $g(x)$ is even.", not "$=g(x)$ which is even."
For (b) you could say how you know/why you think $\sin(x)$ is not injective.
(c) I think is fine, although you could give more detail about the domain and range of the functions throughout your working.