Step 1: Understand the question.
The piecewise function uses $\dfrac{x^2-4}{\sqrt{2-x}}$ for $x<2$, the value $a$ at $x=2$, and $\log(x-2)$ for $x>2$. We test continuity at $x=2$.
Step 2: Left-hand limit.
For $x<2$, factor $x^2-4=(x-2)(x+2)$ and note $x-2=-(2-x)$:
\[ \frac{(x-2)(x+2)}{\sqrt{2-x}}=-(2-x)^{1/2}(x+2). \]
Step 3: Evaluate the left limit.
As $x\to2^-$, $(2-x)^{1/2}\to0$, so the left-hand limit is $0$.
Step 4: Right-hand limit.
As $x\to2^+$, $x-2\to0^+$, so $\log(x-2)\to-\infty$. The right-hand limit is not finite.
Step 5: Two-sided continuity is impossible.
Since the right limit is $-\infty$ while the left limit is $0$, $f$ can never be (fully) continuous at $x=2$, whatever $a$ is.
Step 6: Check one-sided continuity.
If $a=0$, then $f(2)=0$ equals the left-hand limit, so $f$ is left continuous at $x=2$ when $a=0$.
\[ \boxed{\text{f is left continuous at }x=2\text{ when }a=0} \]