If $y^2$ is defined by (1) and
$$\epsilon = \frac{168}{97},x= \frac{1168561}{2916},z= \frac{121}{8281},\Delta = \frac{1}{1000},K=
\frac{10201}{9604},$$
then the difference between the right-hand side of inequality (2) and its left-hand side is $-2.5248\ldots<0$, so that (2) fails to hold.
(What might be interesting, though, is that, for $y^2,\epsilon,x,z,\Delta,K$ as above, the values of the right-hand side and left-hand side are relatively close to each other: $2635.1\ldots$ and $2637.6\ldots$.)
Also, if $y^2$ is defined by (1), $y>0$, $V=y^{-1}\left(2\Delta y^{-1} + \sqrt{2x} + \sqrt{2\delta}\right)$ and
$$K= \frac{602}{61},x= \frac{7936}{57},\delta = \frac{34618}{41},\epsilon = \frac{6}{73},\Delta =
\frac{37534}{29},$$
then the ratio of the left-hand side of inequality
$$\epsilon V \le K^{-\frac 12} \tag{*}$$
to its right-hand side is $1.2496\ldots>1$, so that (*) fails to hold, too.
Application of Holder's inequality
Notice that the estimate on the Green's function means that
$$ \int |G(P,Q)|^\alpha ~dQ $$
is bounded whenever $\alpha < \frac{n}{n-2}$ (and the bound can be taken to be uniform; that is independent of $P$).
By Holder's inequality, we have
$$ \int G(P,Q) F(Q) ~dQ \leq \|G(P,-)\|_{L^\alpha} \|F\|_{L^\beta} $$
when $\beta^{-1} = 1 - \alpha^{-1}$. To ensure that the first term is bounded, we need $\alpha < \frac{n}{n-2}$ as described above; this means that $\beta^{-1} < 1 - \frac{n-2}{n} = \frac{2}{n}$.
So applying to equation (2) (noting that $h$ and $f$ are smooth, bounded, and can be discarded), we see that if $\varphi$ is such that both $\varphi\in L^\beta$ and $\varphi^{q-1} \in L^{\beta'}$ with $\beta, \beta' > \frac{n}{2}$, then we can apply the chain of reasoning above to conclude that $\varphi$ is uniformly bounded.
By assumption $q - 1 > 1$; if $\varphi^{q-1}\in L^{\beta'}$ then $\varphi\in L^{(q-1)\beta'}$. This means that we only need:
if $\varphi^{q-1} \in L^{\beta'}$ with $\beta' > \frac{n}2$, then the equation will guarantee that $\varphi\in L^\infty$.
This final condition can be re-written as $\varphi \in L^\beta$ for some $\beta > \frac{n}{2} (q-1)$. So you just need to argue somehow that $\varphi$ can be taken to have this degree of integrability.
Iteration process
Pretty much your intuition is correct. The idea is that solving equation (1), if you know that $\varphi$ is a priori in $L^\beta$, with $\beta < \frac{n}2(q-1)$, then the Sobolev inequality will tell you that
$$ \varphi \in L^{\gamma}, \quad \gamma = \frac{n \beta }{n(q-1) - 2\beta} $$
Our hope is that starting with $\varphi\in L^\beta$, we can improve it to some $L^\gamma$ for $\gamma > \beta$. And repeating this improvement should eventually get us above the threshold $\frac{n}2(q-1)$.
Examining this $\gamma$, we see that (under still the assumption that $\beta < \frac{n}2(q-1)$)
$$ \gamma > \beta \iff \frac{n}{n(q-1) - 2\beta} > 1 \iff 2 + \frac{2\beta}{n} > q $$
In particular: if we start the iteration process with the first step such that $\gamma > \beta$, then for all subsequent steps we will continue to get improvements from the process.
For the first step, we have $\beta = \frac{2n}{n-2}$ by Sobolev embedding since we pre-supposed $\varphi\in H^1$. So if we know that
$$ q < 2 + \frac{2\beta}{n} = 2 + \frac{4}{n-2} = \frac{2n}{n-2} $$
we can guarantee that during the iteration process, starting from initial knowledge that $\varphi\in H^1$, that at every step we will improve that integrability of $\varphi$.
Best Answer
For the second, rewrite $F(t) \leq \lambda$ as $$t - \lambda \leq \left(\int_{\mathbb{R}} a(s,\,t)\phi(s)\,ds\right)^q,$$ which reduces the problem to showing that $$\int_{\mathbb{R}} a(s,\,t)\phi(s)\,ds \leq (b^q + t)^{1/q}(1-L^p(t))^{1/p} + bL(t).$$ To verify this inequality write the left side as $$\int_{-\infty}^t a(s,\,t)\phi(s)\,ds + \int_t^{\infty}a(s,\,t)\phi(s)\,ds,$$ and apply Holder's inequality to each term. The second term is clearly bounded by $bL(t)$. For the first, use that $\int_{-\infty}^t \phi^p \leq 1-L^p(t)$ (this uses $\int_{\mathbb{R}}\phi^p \leq 1$) and that $\int_{-\infty}^t a^q(s,\,t) = \int_{-\infty}^0 a^q + \int_0^t a^q \leq b^q + t$ (using the definition of $b$ for the first, and the bound of $1$ on $a$ in the relevant interval for the second).