= Solution
Multiplying the desired representation by $b_1b_2$ turns it into the integer equation
$$
a=a_1b_2+a_2b_1+nb_1b_2.
$$
Since $b_1$ and $b_2$ are <coprime integers>, the necessary <modular multiplicative inverses> exist. Choose the unique representatives
$$
a_1\equiv a b_2^{-1}\pmod{b_1},\quad0\leq a_1<b_1,\qquad
a_2\equiv a b_1^{-1}\pmod{b_2},\quad0\leq a_2<b_2.
$$
Then $a-a_1b_2-a_2b_1$ is divisible by both $b_1$ and $b_2$, hence by their product, because they are <coprime>. Defining
$$
n=\frac{a-a_1b_2-a_2b_1}{b_1b_2}
$$
therefore gives an integer and proves existence. Any other representation, reduced modulo $b_1$ and $b_2$, must have the same $a_1$ and $a_2$, since their allowed ranges contain exactly one representative of each residue. The integer equation then forces the same $n$. This proves uniqueness.
For the <prime-power fraction decomposition>, put $d_i=p_i^{n_i}$ and $B_i=b/d_i$. The distinct prime-power factors $d_i$ are pairwise <coprime>, and $B_i$ has a <modular multiplicative inverse> modulo $d_i$. Define
$$
q_i\equiv aB_i^{-1}\pmod{d_i},\qquad0\leq q_i<d_i.
$$
Every term $q_jB_j$ with $j\ne i$ is divisible by $d_i$, while $q_iB_i\equiv a\pmod{d_i}$. Thus $a-\sum_iq_iB_i$ is divisible by all the pairwise coprime $d_i$ and therefore by $b$. With $n=(a-\sum_iq_iB_i)/b$, division by $b$ yields the required representation. Reducing any such representation modulo each $d_i$ proves uniqueness exactly as above. This is the residue-solving mechanism of the <Chinese remainder theorem>.
For $315=9\cdot5\cdot7$, the complementary factors are $35,63,45$. The required <modular inverses> give
$$
q_9=8,\qquad q_5=2,\qquad q_7=5,
$$
because $35\cdot8\equiv1\pmod9$, $63\cdot2\equiv1\pmod5$, and $45\cdot5\equiv1\pmod7$. Their numerator sum is $280+126+225=631=1+2\cdot315$. Thus \b[the unique decomposition] is
$$
\boxed{\frac1{315}=\frac89+\frac25+\frac57-2}.
$$
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