= Solution
For the <First Chern class> identity, use finite-dimensional classifying maps. A <complex line bundle> on a <compact Hausdorff space> embeds in a finite-dimensional <trivial vector bundle>, so it is a <pullback vector bundle> of the <tautological bundle> on some <Complex projective space>. For the two <complex line bundles>, the universal <tensor product of vector bundles> is classified by the map
$$
m:\mathbb{CP}^N\times\mathbb{CP}^M\longrightarrow\mathbb{CP}^{(N+1)(M+1)-1},\qquad ([v],[w])\longmapsto[v\otimes w].
$$
Its restriction to either factor pulls the <tautological bundle> back to the <tautological bundle> on that factor. The degree-two <cohomology> of the product is the <direct sum> of the degree-two <cohomology> of its factors, so these two restrictions determine its <First Chern class>. Pulling back to $X$ proves
$$
\boxed{c_1(L_1\otimes L_2)=c_1(L_1)+c_1(L_2).}
$$
This proves the integral identity, including possible <torsion elements> in the <cohomology> of $X$; an argument using only curvature forms would not detect those <torsion elements>.
The <splitting principle for complex vector bundles> says that for a <vector bundle> $E\to X$ there is an iterated <projective bundle> $p:F(E)\to X$ for which $p^*$ is injective on integral <cohomology>, and $p^*E$ is a <direct sum> of <complex line bundles>. The same assertion holds with rational coefficients, and a common such space can split any finite collection of <vector bundles>. The injectivity follows at each stage from the <projective bundle formula for complex vector bundles>; a <Hermitian metric> splits the resulting filtration into a <direct sum>.
Write $x_i=c_1(L_i)$ for the <Chern roots> of the pulled-back <vector bundle>. Define the degree-$2j$ component of its <Chern character> by
$$
\operatorname{ch}_0(E)=\operatorname{rank}E,\qquad
\operatorname{ch}_j(E)=\frac{1}{j!}\sum_i x_i^j,\qquad
\operatorname{ch}(E)=\sum_{j\geq0}\operatorname{ch}_j(E).
$$
The <Fundamental theorem of symmetric polynomials> expresses each <power-sum symmetric polynomial> $\sum_i x_i^j$ uniquely as an integral polynomial in the <elementary symmetric polynomials>, which are the <Chern classes> of $E$. Thus the definition makes sense on $X$ itself, without choosing actual <Chern roots> there. On a <compact Hausdorff space>, each <vector bundle> is classified by a finite-dimensional <Grassmannian>; its high-degree <Chern character> components therefore vanish. Hence this defines a class in the direct sum $H^{2*}(X;\mathbb Q)$ even if $X$ has no finite dimension. On more general spaces an infinite <Chern character> is instead interpreted in the product of the even-degree <cohomology> groups.
On a common splitting space, the <Chern roots> of a <direct sum> are the combined root lists, whereas the <First Chern class> identity makes the <Chern roots> of a <tensor product of vector bundles> the pairwise sums $x_i+y_j$. Consequently
$$
\begin{aligned}
\operatorname{ch}(E\oplus F)&=\sum_i e^{x_i}+\sum_j e^{y_j},\\
\operatorname{ch}(E\otimes F)&=\sum_{i,j}e^{x_i+y_j}
=\left(\sum_i e^{x_i}\right)\left(\sum_j e^{y_j}\right).
\end{aligned}
$$
The injectivity in the <splitting principle for complex vector bundles> descends both identities to $X$. Additivity extends the <Chern character> to the <Grothendieck group> by $\operatorname{ch}([E]-[F])=\operatorname{ch}(E)-\operatorname{ch}(F)$. Multiplicativity and $\operatorname{ch}(1)=1$ then show that \b[the <Chern character> is a unital <ring homomorphism>].
For <Complex K-theory of complex projective space>, let $\gamma$ be the <tautological bundle>, put $t=1-[\gamma]$, and set $h=-c_1(\gamma)$. The <cohomology ring of complex projective space> is $\mathbb Z[h]/(h^{n+1})$. The <CW filtration> gives a <cofibration>
$$
\mathbb{CP}^{n-1}\longrightarrow\mathbb{CP}^n\longrightarrow S^{2n}.
$$
The <K-theory six-term exact sequence> and <Complex K-theory of a sphere> give, inductively,
$$
K^{-1}(\mathbb{CP}^n)=0,\qquad
0\longrightarrow\mathbb Z\{b_n\}\longrightarrow K^0(\mathbb{CP}^n)
\longrightarrow K^0(\mathbb{CP}^{n-1})\longrightarrow0.
$$
Here $b_n$ is the image of a <Bott element> from $S^{2n}$, normalized so that $\operatorname{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of <Bott periodicity>: the <Chern character> of the degree-two <Bott element> is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the <abelian group> $K^0(\mathbb{CP}^n)$ is a <free abelian group> of rank $n+1$.
Assume inductively that $1,t,\ldots,t^{n-1}$ form an integral basis on $\mathbb{CP}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on $\mathbb{CP}^n$. Since
$$
\operatorname{ch}(t)=1-e^{-h}=h-\frac{h^2}{2!}+\frac{h^3}{3!}-\cdots,
$$
the <Chern characters> of this basis are <linearly independent> over $\mathbb Q$: their first nonzero degrees are $0,2,\ldots,2n$. Thus the <Chern character> is injective on $K^0(\mathbb{CP}^n)$. By induction $t^n$ restricts to zero on $\mathbb{CP}^{n-1}$, so $t^n=a b_n$ for an integer $a$. Comparing <Chern characters> gives $a=1$. Also $\operatorname{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:
$$
\boxed{K^0(\mathbb{CP}^n)\cong\mathbb Z[t]/(t^{n+1}),\qquad t=1-[\gamma].}
$$
The induction begins with $\mathbb{CP}^0$ a point and $t=0$.
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