The three tasks are interesting. Unfortunately, Euler in “Observationes analyticae variae de combinationibus”, Commentarii academiae scientiarum Petropolitanae, 13, 1751, pp. 64–93. (https://arxiv.org/pdf/0711.3656.pdf is an English translation of E158) and his later works, Caley, Sylvester, Macmahon, Hardi, Ramanujan, Rademacher, … did not leave an opportunity to make a discovery in these three and a few related problems (but not all of this kind).
Terminology: a partition of a positive integer is its representation as a sum of positive integers not considering the order of the summands - parts; compositions are partitions in which the order of occurrences of parts is essential. In general, the number $n$, and $r$ parts are involved. A multiset is a set with possibly repeated elements. Numbers of repetitions are multiplicities. We shall write the multiset $\{1,2,2,8\}$ as $\{1^1,2^2,8^1\}$, where “powers” are multiplicities.
The number of partitions of 13 in any numbers of parts $\{1, 2, 8\}$ can be obtained as a coefficient at $x^{13}$ in the expansion of $\frac{1}{(1-x)(1-x^2)(1-x^8)}$. If this looks strange, then recollect that $\frac{1}{1-x}$ is the infinite geometric series $1+x+x^2+x^3+x^4+\cdots$. Similarly, $\frac{1}{1-x^2}$ is $1+x^2+x^4+x^6+x^8+\cdots$, and $\frac{1}{1-x^8}$ is $1+x^8+x^{16}+x^{24}+x^{32}+\cdots$. When three infinite series are multiplied, the coefficients at the lower powers of $x$ are not affected by the higher powers – stabilize. Our product is $1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+6x^8+6x^9+8x^{10}+8x^{11}+10x^{12}+10x^{13}+12x^{14}+\dots$
So, $3=1+1+1=1+2$ yields two partitions using allowed parts 1, 2, 8. This is the coefficient 2 at the term of $x^3$. But $5=1+1+1+1+1=1+1+1+2=1+2+2$ yields three partitions, notice $3x^5$. 13 yields 10 partitions, notice $10x^{13}$, written for briefness as multisets $\{1^{13}\}$, $\{1^{11},2^1\}$, $\{1^9,2^2\}$, $\{1^7,2^3\}$, $\{1^5,2^4\}$, $\{1^3,2^5\}$, $\{1^1,2^6\}$, $\{1^{5},8^1\}$, $\{1^3,2^1,8^1\}$, $\{1^1,2^2,8^1\}$. The number of unique permutations of multiset $\{p_1^{m_1},\dots,p_k^{m_k}\}$ elements is $\frac{(m_1+\cdots+m_k)!}{m_1!\cdots m_k!}$.
We get the sum of 10 numbers $1+12+55+120+[\frac{(5+4)!}{5!4!}=7\cdot 2 \cdot 9=126]+56+7+6+20+12=415$. This coincides with Gerry's evaluation and supports original result. Euler discusses that sometimes expansions become rather complex and for a few important cases derives recurrent relationships between the coefficients in front of powers. Asymptotic formulas from Hardi, Ramanujan, Rademacher evaluate the number of unrestricted partitions of $n$. Best Regards, Valerii Salov.