(After this answer was posted, a duplicate was noted. This answer is equivalent to the answers by Paganini and Franklin in that earlier post, but with more detail.)
As stated in the question:
$K^+$ has $(I,I_3) = (1/2,1/2)$
$\pi^+$ has $(I,I_3)=(1,1)$
$\pi^0$ has $(I,I_3)=(1,0)$
Therefore, $\pi^+\pi^0$ must have either $I=0$, $1$, or $2$.
The case $I=0$ is not consistent with $I_3=1+0=1$.
To rule out $I=1$, consider the ways to get $I_3=1$ by combining $I_3=1$ ($\pi^+$) with $I_3=0$ ($\pi^0$). The possibilities with definite values of $I$ are:
Case 1: $1\otimes 0 - 0\otimes 1$, which has $I=1$.
Case 2: $1\otimes 0 + 0\otimes 1$, which has $I=2$,
Both cases have $I_3=1$. The $I$ values can be inferred from the fact that applying the $I_3$-raising operator gives zero ($1\otimes 1 - 1\otimes 1$) in case 1 and gives a non-zero result ($1\otimes 1$) in case 2. This shows that if the final two-pion state had $I=1$, then it would need to be antisymmetric in the isospin labels. But pions are bosons, and the total angular momentum is zero (because the initial kaon has spin zero). So, since the pions are distinguished from each other only by their isospin, the state must be symmetric with respect to the isospin labels. This rules out the case $I=1$. (See appendix 1 for more detail.)
The only remaining possibility is that $\pi^+\pi^0$ has $I=2$. Therefore, the process $K^+\rightarrow \pi^+\pi^0$ would have $\Delta I = 3/2$, as the textbook claimed — assuming the textbook was referring to the change in total isospin $I$, not the component $I_3$.
Appendix 1
Here is a more detailed explanation of why the two-pion state must be symmetric with respect to the isospin labels. (This was used above to rule out $I=1$.) For simplicity, work in a non-relativistic model in which $a(x)$ and $b(x)$ are the annihilation operators for $\pi^+$ and $\pi^0$, respectively, at a spatial point $x$, and $a^*(x)$ and $b^*(x)$ are the corresponding creation operators. Consider the two states
$$
|\pm\rangle = \int d^3x\,d^3y\ f(x,y)\big(a^*(x)b^*(y)\pm b^*(x)a^*(y)\big)
|0\rangle
$$
where $|0\rangle$ is the vacuum state. The signs $+$ and $-$ correspond to $I=2$ and $I=1$, respectively, as explained before. The goal here is to show that the antisymmetric case ($-$) cannot have zero angular momentum.
Pions are bosons, so the operators $a$ and $b$ commute with each other. Use this to get
$$
|\pm\rangle = \int d^3x\,d^3y\ \big(f(x,y)\pm f(y,x)\big)a^*(x)b^*(y)
|0\rangle.
$$
Working in the center-of-mass frame, the total momentum must be zero, so the wavefunction $f(x,y)$ can only depend on the difference $x-y$. Then, to have zero angular momentum about the origin, the wavefunction must be invariant under rotations about the origin, so it can only depend on $|x-y|$. This implies $f(x,y)-f(y,x)=0$, because $|x-y|=|y-x|$. This implies that $|-\rangle$ must be the zero vector, which does not represent a physical state. Therefore, as claimed, the angular momentum cannot be zero in the antisymmetric case.
Appendix 2
Everything in this answer assumes that the "kaon" in question is the charged kaon of lowest mass, usually denoted $K^+$. According to the Particle Data Group (http://pdg.lbl.gov/2018/tables/contents_tables_mesons.html), the $K^+$ has total angular momentum zero. More-massive and shorter-lived particles/resonances with the same valence quark content ($u\bar s$) can also be produced, with non-zero total angular momentum. Whether the non-zero total angular momentum is due to orbital or spin angular momentum (or both) is an interesting quesiton that I won't try to answer here, but see https://arxiv.org/abs/0803.2775 for a related question applied to the proton.
You might profit from this old talk by Nichitiu. The "theoretical" ratio of coupling widths to the photon is, in fact, 9 to 1.
The respective flavor wavefunctions of light vector mesons are, schematically, skipping γs, color, spin, etc...
$$
\rho^0\sim \frac{\bar u u -\bar dd}{\sqrt{2}}; ~~\omega\sim \frac{\bar u u +\bar dd}{\sqrt{2}};~~ \phi \sim \bar s s.
$$
The VDM EM couplings to the photon, then, are the overlaps of the EM current with the above wavefunctions, respectively: effective charges. Squared for the width, they then lead to the ratios
$$
\Gamma_{ee} (\rho)~: \Gamma_{ee} (\omega)~:\Gamma_{ee} (\phi) \\ = \left ({2/3+1/3\over\sqrt{2}}\right )^2~: \left ({2/3-1/3\over\sqrt{2}}\right )^2~: (-1/3)^2 \\ = 9:1:2 ~.
$$
I'm not sure what your second question might mean: the point of the generalized Pauli principle is that if you wrote a wave-function with the wrong (anti)symmetry, it would vanish. But you see manifestly and explicitly that the two wavefunctions above, contrasting the isovector to the isosinglet, need not vanish... how could they? (The analogous thinking would, by contrast lead to zeros in baryon wavefunctions.)
Best Answer
In a travesty of overlapping historical notations, we have both strong isospin, under which the neutron and proton are the two projections of the nucleon and are raised and lowered by the pion, and weak isospin, in which the left-handed parts of the $(u,d)$, $(e,\nu_e)$, $(c,s)$, $(\mu, \nu_\mu)$, $(t,b)$, and $(\tau,\nu_\tau)$ doublets are raised and lowered by the $W^\pm$ bosons. The six right-handed quarks and the six right-handed leptons are weak isospin singlets.
It is possible in principle (I think) for a particle like the $\pi^0$ to be the neutral member of a strong isospin triplet, but a singlet under weak isospin.
However, that doesn't seem to be the case for the $\pi^0$. The "lowering operator" vertex $\pi^\pm \leftrightarrow W^\pm\pi^0$ actually does exist, as evidenced by the existence of the decay mode $\pi^+\to \pi^0 e^+ \nu_e$, which has branching ratio $10^{-8}$. You have a fresh comment suggesting that your textbook makes this claim incorrectly.