= Solution
If $\phi$ is identically zero there are no exceptions. Otherwise choose $x_0$ with $\phi(x_0)>0$ and put $I_+=\{x\ge x_0:\phi'(x)>\phi(x)^2\}$. Monotonicity implies $\phi(x)>0$ thereafter. On this set, $1<\phi'/\phi^2$, whence
$$
|I_+|\le\int_{I_+}\frac{\phi'}{\phi^2}\,dx\le\int_{x_0}^\infty\frac{\phi'}{\phi^2}\,dx
=\frac1{\phi(x_0)}-\lim_{X\to\infty}\frac1{\phi(X)}\le\frac1{\phi(x_0)}.
$$
Adding the finite initial interval $[0,x_0]$ proves
$$
\boxed{\phi'(x)\le\phi(x)^2\quad\text{outside a set of finite measure}.}
$$
This <derivative growth outside a finite-measure set> controls exceptional radii in Nevanlinna estimates. For example let $T_s$ be the smooth spherical characteristic and $A(r)=rT_s'(r)$ its increasing area-counting function. Apply the lemma to $T_s+1$ and $A+1$: outside the union of two finite-measure sets,
$$
A(r)\le r(T_s(r)+1)^2,\qquad A'(r)\le(A(r)+1)^2.
$$
Their logarithms are consequently $O(\log r+\log(T_s+1))$. Such <derivative>/area bounds in estimates from the <Poisson-Jensen formula> give the <Nevanlinna logarithmic derivative lemma>, $m(r,f'/f)=O(\log^+T_f+\log r)$ outside a finite-length set. The second main theorem uses that estimate as its error term. The elementary growth lemma controls that error and its exceptional set; it is not itself a substitute for the analytic estimates.
We use the truncated second main theorem stated in the preceding solution, together with the fact that a transcendental <meromorphic function> has $T_f(r)/\log r\to\infty$. Here is a justification of the latter fact. If there are infinitely many <poles>, fixing any arbitrarily large finite <pole> count gives $N(r;\infty)\ge A\log r-O_A(1)$. If there are only finitely many <poles>, multiply by a <polynomial> $p$ removing them to obtain a transcendental entire function $h=pf$. Its <Taylor series> has nonzero coefficients of arbitrarily high degrees $n$. Cauchy's coefficient estimate gives $\log M_h(r/2)\ge n\log r-O_n(1)$; the <subharmonic function> Poisson estimate gives $\log M_h(r/2)\le3m(r,h)$. Since $T_h\le T_f+O(\log r)$, letting $n$ be arbitrarily large proves the claimed ratio limit. Thus the second-theorem error is $o(T_f)$ along large nonexceptional radii.
Let $N_{\rm simple}(r;a)$ count just the simple $a$-points. A multiple $a$-point of degree $m\ge2$ contributes one to the truncated count and at least two to the full count; a simple point contributes one to each. Hence, with the same positive logarithmic weights for large radii,
$$
\overline N(r;a)\le\tfrac12N(r;a)+\tfrac12N_{\rm simple}(r;a).
$$
If each of the five values had only finitely many simple preimages, each simple counting function would be $O(\log r)$. Summing and using the first main theorem would give $\sum_{j=1}^5\overline N(r;a_j)\le(5/2)T_f+O(\log r)$. But the second theorem requires $3T_f\le\sum\overline N+o(T_f)$, a contradiction. This proves the stronger result <five values force infinitely many simple preimages>:
$$
\boxed{\text{At least one of the five values has infinitely many simple preimages.}}
$$
In particular \b[more than one simple zero is necessary], but \b[more than one target with simple zeros is not necessary].
For the last distinction, take a nonsingular <Weierstrass elliptic function> scaled so that $\wp'^2=4\wp(\wp-1)(\wp+1)$; it has branch values $-1,0,1,\infty$. These are the <four totally ramified Weierstrass values>: finite branch preimages have local degree two, and all <poles> are double. Set
$$
f(z)=\frac1{\wp(z)-2},\qquad
\{a_1,\ldots,a_5\}=\{-\tfrac13,-\tfrac12,-1,0,1\}.
$$
The first four targets are the images of those branch values under the <Möbius transformation>. Their preimages remain double; at a lattice <pole>, for example, $f(z)\sim(z-z_*)^2$. The last target corresponds to $\wp=3$, where $\wp'^2=96\ne0$, so every such preimage is simple. The degree-two torus map attains that regular value, and periodicity produces infinitely many preimages in the plane. This nonconstant <elliptic function> is transcendental because it has infinitely many <poles>. Exactly one of the five selected finite targets therefore has simple preimages.
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