Finding the value of $k$ for an Uniform Distribution defined on $(-k,k)$ 
If $X$ be an uniform distribution defined on $(-k,k)$, then the value of $k$ for so that :
$$P(|X|<1) = P(|X|>2)$$

I began by defining the $p.d.f$ of the Uniform function namely:
$$ f(x) = \frac{1}{2k}$$
Then we can re-write the LHS of the probability as :
$$ P(|X|<1) = P(-1<X<1)$$
And the RHS :
$$ P(|X|>2) = 1-P(|X|<2) \ = 1-P(-2<X<2) $$
Now solving the probability by $integration$ we get:
$$\int_{-1}^{1} \frac{1}{2k} dx \ = 1  -  \int_{-2}^{2}\frac{1}{2k}dx$$
$$\frac{1}{k} = 1  -   \frac{2}{k}$$
And therefore
$$k=3$$
Is my method and answer correct?

Your result is correct, but the method has a small flaw.
Your density should be
$$
f(x) = \frac{1}{2k} \cdot I_{(-k,k)} (x)
$$
This didn't change anything, since both your integration areas lie in $(-3,3)$, but in general this could fail.
