An example member of $C[-1, 1]$ is $f_1(x) = (x+1)(x-1),\quad -1 \leq x \leq 1$, it's also an example of a function that satisfies $f_1(-1) = f_1(1)$.
$f_2(x) = (x+1)(2-x), \quad -1 \leq x \leq 1$, is an example of a member that has $f_2(-1) = 0$ or $f_2(1) = 0$ because $f_2(-1) = 0$ but $f_2(1) = 2 \neq 0$.
Looking at examples always helps to understand and also can provide counterexamples when you're proving something false. When it's true, you ultimately have to be able to prove it generally, like with (a).
I suppose I should mention, just in case it introduces confusion that $f_1$ also satisfies the (b) condition as it is zero at both extremes. Logical binary "or" only fails if both arguments fail.
$f_3 = x^2 + 2$ satisfies (a) but not (b) as $f_3(-1) = f_3(1) = 3 \neq 0$.
For (a), if $f(-1) = f(1)$ then $kf(-1) = kf(1)$.
Also, if $g(-1) = g(1)$ then $f(-1) + g(-1) = f(1) + g(1)$.
With problems like this, just grab a piece of paper and a pen and jump in. You find once you start writing things out they'll fall into place. Don't just try and think it out.
For (b) the "or" is the clue, try $f$ being zero only on the left and $g$ being zero only on the right. What happens when you add them together? Problems with function spaces like this usually involve point wise operations so you're really just adding numbers together and seeing what happens.
Answer to (b): If $f(-1) = 0$, $f(1) = a \neq 0$, $g(-1) = b \neq 0$, $g(1) = 0$, these both satisfy the requirements for at least one of these values to be zero.
(f+g)(-1) &= f(-1) + g(-1) = 0 + b = b \neq 0 \\
(f+g)(1) &= f(1) + g(1) = a + 0 = a \neq 0
And so $f+g$ does not satisfy the requirements as both values are non zero. This is a counterexample. We only need to show one where it's not a closed subset, so it's not a subspace.