a) We notice that $(b_n)_{n \ge 1}$ is non-decreasing. Also, the sequence $(a_n)_{n \ge 1}$ is non-increasing, since $0 \le a_{mn} \le a_n - a_{n+1}$. Moreover,
$$\sum_{k=1}^{n} a_{mk} \le a_1 - a_{n+1} \le a_1.$$

Using the monotonicity of $(a_n)$ and $(b_n)$, we have:
$$b_n \le b_{mn} = \sum_{k=1}^{mn} a_k = \sum_{i=1}^{m-1} a_i + \sum_{k=1}^{n} a_{mk} + \sum_{k=1}^{n-1} \sum_{j=1}^{m-1} a_{mk+j}.$$
By monotonicity,
$$\sum_{k=1}^{n-1} \sum_{j=1}^{m-1} a_{mk+j} \le (m-1) \sum_{k=1}^{n-1} a_{mk} \le (m-1)a_1,$$
hence,
$$b_n \le \sum_{i=1}^{m-1} a_i + \sum_{k=1}^{n} a_{mk} + (m-1)a_1 \le \sum_{i=1}^{m-1} a_i + ma_1,$$

b) We first prove that the sequence $d_n = \sum_{k=1}^{n} k a_k$ is bounded above. Clearly, $(d_n)_{n \ge 1}$ is non-decreasing. Moreover,
$$\sum_{k=1}^{n} k a_{mk} \le \sum_{k=1}^{n} k(a_k - a_{k+1}) = a_1 + \sum_{k=2}^{n} (k - (k-1))a_k - n a_{n+1} \le b_n,$$
so the sequence $\left( \sum_{k=1}^{n} k a_{mk} \right)_{n \ge 1}$ is bounded above.
On the other hand,
$$\begin{aligned} d_n \le d_{mn} &= m \sum_{k=1}^{n} k a_{mk} + \sum_{k=1}^{m-1} k a_k + \sum_{j=1}^{m-1} \sum_{k=1}^{n-1} (mk + j)a_{mk+j} \\ &\le m b_n + d_{m-1} + (m-1) \sum_{k=1}^{n-1} m(k+1)a_{mk} \\ &\le m b_n + d_{m-1} + m(m-1)(b_{n-1} + a_1), \end{aligned}$$
hence $(d_n)_{n \ge 1}$ is bounded above.

Again, $(c_n)_{n \ge 1}$ is non-decreasing. Similarly,
$$\begin{aligned} \sum_{k=1}^{n} k^2 a_{mk} &\le \sum_{k=1}^{n} k^2 (a_k - a_{k+1}) = a_1 + \sum_{k=2}^{n} (k^2 - (k-1)^2)a_k - n^2 a_{n+1} \\ &\le a_1 + \sum_{k=2}^{n} 2k a_k \le 2 d_n, \end{aligned}$$
so the sequence $\left( \sum_{k=1}^{n} k^2 a_{mk} \right)_{n \ge 1}$ is bounded above.
Finally,
$$\begin{aligned} c_n \le c_{mn} &= m^2 \sum_{k=1}^{n} k^2 a_{mk} + c_{m-1} + \sum_{j=1}^{m-1} \sum_{k=1}^{n-1} (mk + j)^2 a_{mk+j} \\ &\le 2m^2 d_n + c_{m-1} + m^2(m-1) \left( \sum_{k=1}^{n-1} k^2 a_{mk} + 2 \sum_{k=1}^{n-1} k a_{mk} + \sum_{k=1}^{n-1} a_{mk} \right), \end{aligned}$$
which is a finite sum of bounded above sequences, hence $(c_n)_{n \ge 1}$ is bounded above.