= Solution
There are \b[three independent nonfactorized <momentum integrals>], namely a <tadpole integral>, the three-line <sunset diagram> integral, and the <five-line banana integral>. The six displayed <Feynman diagrams> involve \b[six complete integral expressions]:
$$
\boxed{T,\quad T^2,\quad B_3(P),\quad T B_3(P),\quad T^2B_3(P),\quad B_5(P).}
$$
The distinction between independent building blocks and their products removes an ambiguity in counting “different momentum integrals”. For the strict divergent-<wave-function renormalization> interpretation of <3/c/ii>, the graphs in <3/c> reduce to $T,T^2,B_5$, comprising \b[three full integral expressions and two independent nonfactorized building blocks].
Make a <Wick rotation> and use $m>0$ to separate <ultraviolet divergences> from <infrared divergences>. Write $D_E(k)=1/(k^2+m^2)$, and let $T,B_r(P)$ be the Euclidean versions of the integrals above, with their harmless overall factors of $i$ removed. A spherical <ultraviolet cutoff> gives
$$
T_\Lambda=\frac1{2\pi^2}\int_0^\Lambda\frac{k^2\,dk}{k^2+m^2}
=\frac1{2\pi^2}\left[\Lambda-m\arctan(\Lambda/m)\right]
=\frac{\Lambda}{2\pi^2}-\frac{m}{4\pi}+O(\Lambda^{-1}).
$$
Thus the one-vertex quartic graph is linearly divergent, and the one-vertex sextic graph is quadratically divergent as $T_\Lambda^2$.
For the other two building blocks, the spatial <Fourier transform> is especially transparent. The three-dimensional Euclidean <scalar propagator> is
$$
G(x)=\int\frac{d^3k}{(2\pi)^3}\frac{e^{ik\cdot x}}{k^2+m^2}
=\frac{e^{-m|x|}}{4\pi|x|}.
$$
The <convolution theorem> converts the $r$ parallel lines into $G(x)^r$. Angular integration gives, with a short-distance cutoff $a=\Lambda^{-1}$,
$$
B_r^{(a)}(P)=\frac{4\pi}{(4\pi)^r}\int_a^\infty
\rho^{2-r}e^{-rm\rho}\frac{\sin(P\rho)}{P\rho}\,d\rho,
\qquad P=|P_E|.
$$
This is a different convenient <cutoff regularization> for the connected block; power-divergent coefficients depend on the cutoff convention, whereas the conclusions about powers and the presence of a kinetic logarithm do not. The <sinc function> is understood to be $1$ at $P=0$.
For $r=3$, the integrand starts as $\rho^{-1}+O(1)$, so
$$
B_3^{(a)}(P)=\frac1{16\pi^2}\log\frac1{ma}+O(1).
$$
Its divergent part is independent of external <momentum>. In particular,
$$
B_3(P)-B_3(0)=\frac1{16\pi^2}\int_0^\infty\frac{e^{-3m\rho}}\rho
\left[\frac{\sin(P\rho)}{P\rho}-1\right]d\rho
=-\frac{P^2}{864\pi^2m^2}+O(P^4).
$$
The finite difference and its finite derivative show why the pure three-line <sunset diagram> does not need a divergent <wave-function renormalization> in three dimensions.
For $r=5$, the small-distance <Taylor expansion> is
$$
\rho^{-3}e^{-5m\rho}\frac{\sin(P\rho)}{P\rho}
=\rho^{-3}-5m\rho^{-2}
+\left(\frac{25m^2}{2}-\frac{P^2}{6}\right)\rho^{-1}+O(1).
$$
Consequently
$$
\boxed{B_5^{(a)}(P)\big|_{\mathrm{div}}
=\frac1{256\pi^4}\left[\frac1{2a^2}-\frac{5m}{a}
+\left(\frac{25m^2}{2}-\frac{P^2}{6}\right)\log\frac1{ma}\right].}
$$
In particular, the nonzero $P^2\log\Lambda$ term requires a genuine divergent <wave-function renormalization>, in addition to the <mass counterterm>. The coefficients here refer to the displayed position-space cutoff and have not included <Feynman-diagram symmetry factors> or <bare couplings>.
The comparison with the <superficial degree of divergence> is
$$
\begin{array}{c|c|c|c}
\text{vertices and integral}&(n_4,n_6)&\omega&\text{leading unrenormalized growth}\\\hline
4:\ T&(1,0)&1&\Lambda\\
6:\ T^2&(0,1)&2&\Lambda^2\\
4,4:\ B_3&(2,0)&0&\log\Lambda\\
4,6:\ TB_3&(1,1)&1&\Lambda\log\Lambda\\
6,6:\ T^2B_3&(0,2)&2&\Lambda^2\log\Lambda\\
6,6:\ B_5&(0,2)&2&\Lambda^2
\end{array}
$$
Thus the actual leading power agrees with $\omega$ in every case. The additional logarithms in products come from <ultraviolet subdivergences>, which uniform <power counting> alone does not resolve.
For $TB_3$ and $T^2B_3$, differentiation with respect to $P^2$ removes the logarithmic divergence of $B_3$, but leaves divergent factors $T$ or $T^2$ multiplying a finite, nonpolynomial function of $P^2$. These are subdivergent contributions, not new nonlocal <counterterms>: subtracting each <tadpole diagram> subdivergence and the momentum-independent $B_3$ subdivergence removes them. For example, in a factorized subtraction scheme $TB_3\mapsto(T-T_{\mathrm{div}})(B_3-B_{3,\mathrm{div}})$, and likewise for $T^2B_3$. The remaining $B_5$ kinetic logarithm explains the strict interpretation in <3/c/ii>. A <dimensional regularization> scheme can suppress power divergences; it does not invalidate this cutoff-based comparison with $\omega$.
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