As Iosif said, in general the system you specified does not admit a solution. Here we will give a more pedestrian argument using only comparisons.

## Monotonicity

**Claim**: if a solution exists, and $f(1) > 0$, then the function is monotonically decreasing; if $f(1) < 0$, then the function is monotonically increasing.

*Proof*: we will focus on the positive case. The negative case is similar.
Let $\zeta = \frac{4a}{4a + m(m+1)} \in (0,1)$ (if $a\in (0,\frac12)$). The second derivative test shows that $f$ cannot have a local maximum with $f(r) > \zeta f(1)$ or a local minimum with $f(r) < \zeta f(1)$. This immediately implies monotonicity in light of $f(1) > 0 = \lim f(r)$.

*More details on monotonicity proof*: Given $f(1) > f(\infty) = 0$. Suppose $f$ were not monotonic. Then there exists $r_m, r_M$ with $1 < r_m < r_M < \infty$ such that $f(r_m) < f(r_M)$.

I claim that $f(r_M) > 0$ and $f(r_m) < f(1)$. Suppose not: if $f(r_M) \leq 0$ then $f(r_m) < 0$ and $f$ would have a negative local minimum, in contradiction to the second derivative test. If $f(r_m) \geq f(1)$ then $f$ would have a local maximum with value $> f(1)$, against in contradiction to the second derivative test.

Therefore there must exist a local minimum $s_m\in (1,r_M)$ with $f(s_m) \leq f(r_m)$ and a local maximum $s_M\in (r_m,\infty)$ with $f(s_M) \geq f(r_M)$.

We have therefore established
$$ \zeta f(1) \leq f(s_m) < f(s_M) \leq \zeta f(1) $$
which is a contradiction.

In particular, we must have $f' \leq 0$ on $[1,\infty)$.

## Comparison

We have then

$$ f'(1) - f'(r) = -\int_1^r f''(s) ~ds = \int_1^r \frac{4a}{s^2 - 2as} f(1) - \frac{4a + m(m+1)}{s^2 - 2a s} f(s) + \frac{2(s-a)}{s^2 - 2as} f'(s) ~ds $$

Since $f'$ is signed, we know that it is absolutely integrable on $[1,\infty)$. Since $f$ is monotonic (and hence bounded) the second integrand is also absolutely integrable. We conclude then that $\lim_{r\to\infty} f'(r)$ exists.

Since $\lim_{r\to\infty} f(r) = 0$, we must have also $\lim_{r\to\infty} f'(r) = 0$.

But now writing $f'(r) = - \int_r^\infty f''(s) ~ds$ using the above formula, we see that asymptotically $|f'(r)| \sim \frac1r$ (coming from the $4a f(1)$ term if it is non-zero; the other two terms can both be bounded by $O(1/r) f(r) = o(1/r)$). But this contradicts the integrability of $f'(r)$.

And hence we have proved:

**Claim**: no solution can exist with $f(1) \neq 0$.

## Uniqueness

When $f(1) = 0$, the same maximum principle argument shows that $f$ must be identically zero. This shows that

**Theorem** The only solution to your system is $f \equiv 0$, with $f(1) = f'(1) = 0$ and $C = 0$.

## Final remark

Heuristically, if you want to look for asymptotically constant solutions to your equation, you probably want it to converge to $\zeta f(1)$ in the limit.