Four, (20 points) Given a real number $a$, solve the inequality for $x$
$$
\sqrt{3 x-5 a}-\sqrt{x-a}>1 \text {. }
$$
From the original inequality, we have
$$
\sqrt{3 x-5 a}>\sqrt{x-a}+1\left(x \geqslant a \text {, and } x \geqslant \frac{5}{3} a\right) \text {. }
$$

By squaring both sides, we get
$$
3 x-5 a>x-a+1+2 \sqrt{x-a}(x \geqslant a) \text {. }
$$

Thus, the original inequality is equivalent to
$$
2 x-4 a-1>2 \sqrt{x-a}(x \geqslant a) \text {. }
$$

Let $\sqrt{x-a}=y(y \geqslant 0)$, then $x=a+y^{2}$.
Therefore, equation (1) is equivalent to
$$
2 y^{2}+2 a-4 a-1>2 y(y \geqslant 0),
$$

which simplifies to $\left(y-\frac{1}{2}\right)^{2}>a+\frac{3}{4}(y \geqslant 0)$.
(1) If $a+\frac{3}{4}\sqrt{a+\frac{3}{4}} \text {, }
$$

then $y>\frac{1}{2}+\sqrt{a+\frac{3}{4}}$ or $0 \leqslant y < \frac{1}{2}+\sqrt{a+\frac{3}{4}}$, hence
$$
x>2 a+1+\sqrt{a+\frac{3}{4}}\left(a \geqslant-\frac{1}{2}\right) .
$$
(ii) If $\frac{1}{2}>\sqrt{a+\frac{3}{4}}$, i.e., $-\frac{3}{4} \leqslant a < -\frac{1}{2}$, then
$$
x>2 a+1+\sqrt{a+\frac{3}{4}} \text{ or } a \leqslant x < 2 a+1+\sqrt{a+\frac{3}{4}} \text{; }
$$
When $a \geqslant-\frac{1}{2}$, $x>2 a+1+\sqrt{a+\frac{3}{4}}$.