Among the statements
(S1) : The set $$\{z \in \mathbb{C} - \{-i\} : |z| = 1$$ and $$\frac{z-i}{z+i}$$ is purely real$$\}$$ contains exactly two elements, and
(S2) : The set $$\{z \in \mathbb{C} - \{-1\} : |z| = 1$$ and $$\frac{z-1}{z+1}$$ is purely imaginary$$\}$$ contains infinitely many elements.

We test each statement separately.

Case 1 (S1)
Given $$|z| = 1,\; z \neq -i$$ and $$\frac{z-i}{z+i}$$ is purely real.
For a complex number $$w$$, “purely real’’ means $$w = \overline{w}$$.

Put $$w = \frac{z-i}{z+i}$$.
Taking conjugate and using $$\overline{i} = -i$$ and $$\overline{z}=1/z$$ (because $$|z|=1$$), we get
$$\overline{w}= \frac{\overline{z}+i}{\overline{z}-i}= \frac{\dfrac{1}{z}+i}{\dfrac{1}{z}-i}.$$

Requirement $$w=\overline{w}$$ gives
$$\frac{z-i}{z+i}= \frac{\dfrac{1}{z}+i}{\dfrac{1}{z}-i}.$$ Cross-multiplying, $$(z-i)\Bigl(\frac{1}{z}-i\Bigr)= (z+i)\Bigl(\frac{1}{z}+i\Bigr).\tag{-1}$$

Multiply both sides by $$z$$ to clear the denominator: $$(z-i)(1-iz)= (z+i)(1+iz).\tag{-2}$$

Expand left side:
$$z - iz^2 - i + i^2z = z - iz^2 - i - z = -iz^2 - i.$$
Expand right side:
$$z + iz^2 + i + i^2z = z + iz^2 + i - z = iz^2 + i.$$

Setting the two results equal (from $$(2)$$):
$$-iz^2 - i = iz^2 + i \;\Longrightarrow\; -iz^2 - i - iz^2 - i =0$$
$$\Rightarrow\; -2iz^2 - 2i=0 \;\Longrightarrow\; z^2 + 1 = 0.$$

Hence $$z^2 = -1 \;\Longrightarrow\; z = i \text{ or } z = -i.$$ Because $$z \neq -i$$ (it makes the denominator zero), the only admissible value is $$z = i.$$

Therefore the set in (S1) contains exactly one element, not two. Statement (S1) is incorrect.

Case 2 (S2)
Given $$|z| = 1,\; z \neq -1$$ and $$\frac{z-1}{z+1}$$ is purely imaginary.
“Purely imaginary’’ means $$w + \overline{w} = 0$$ or equivalently $$w = -\overline{w}.$"

Let $$w = $$\frac{z-1}{z+1}$$$$.
Then $$$$\overline{w}= \frac{\overline{z}$$-1}{$$\overline{z}$$+1}= $$\frac{\dfrac{1}{z}$$-1}{\dfrac{1}{z}+1}.$$
Condition $$w = -$$\overline{w}$$$$ gives
$$$$\frac{z-1}{z+1}$$= -$$\frac{\dfrac{1}{z}$$-1}{\dfrac{1}{z}+1}.$$

Clear the denominator as before: $$(z-1)\Bigl($$\frac{1}{z}$$+1\Bigr)= -(z+1)\Bigl($$\frac{1}{z}$$-1\Bigr).\tag{-3}$$ Multiply both sides by $$z$$: $$(z-1)(1+z)= -(z+1)(1-z).\tag{-4}$$

Expand left side:
$$z + z^2 - 1 - z = z^2 -1.$$ Expand right side:
$$-(z+1)(1-z)= -(z - z^2 + 1 - z)= -(-z^2 +1)= z^2 -1.$$

Equality in $$(4)$$ is automatically satisfied for every $$z$$ with $$|z|=1$$; no extra restriction appears.
The only forbidden value is $$z=-1$$ (denominator zero). Therefore every other point on the unit circle works, giving infinitely many solutions.

Thus (S2) is correct.

Conclusion: only statement (S2) is correct. So the correct option is Option C.