Call digits $0, 1, 2, 3, 4$ small and digits $5, 6, 7, 8, 9$ large. Denote the digits of $k$-digit number $n$ from right to left by $(n)_0, (n)_1, \dots, (n)_{k-1}$. Denote by $\Sigma(n)$ the sum of all digits of $n$ and by $l(n)$ the number of large digits of $n$.

a) If $(a)_i$ is small then $(2a)_i = 2(a)_i$ or $(2a)_i = 2(a)_i + 1$ depending on whether $(a)_{i-1}$ is small or large. Similarly if $(a)_i$ is large then $(2a)_i = 2(a)_i - 10$ or $(2a)_i = 2(a)_i - 9$ depending on whether $(a)_{i-1}$ is small or large. In other words, when multiplying by $2$, each large digit necessitates decreasing of the digit at its place by $10$ and increasing of the preceding digit by $1$ in comparison with a small digit. Hence, for every natural number $n$, $\Sigma(2n) = 2\Sigma(n) - 9l(n)$. As $\Sigma(a) = \Sigma(b)$ and $l(a) = l(b)$, we have $\Sigma(2a) = \Sigma(2b)$.

b) If $a = 34$ and $b = 43$ then $\Sigma(3a) = 1 + 0 + 2 = 3$ but $\Sigma(3b) = 1 + 2 + 9 = 12$.

c) Obviously $\Sigma(10a) = \Sigma(10b)$. By part a), on the other hand, $\Sigma(10n) = \Sigma(2 \cdot 5n) = 2\Sigma(5n) - 9l(5n)$, holding for every natural number $n$. Hence $2\Sigma(5a) - 9l(5a) = 2\Sigma(5b) - 9l(5b)$. The digit $(5n)_i$ is large if and only if the digit $(n)_i$ is odd, because a carry during multiplying by $5$ can be at most $4$. Thus $l(5n)$ is the number of odd digits of $n$, implying $l(5a) = l(5b)$. Consequently, $\Sigma(5a) = \Sigma(5b)$.