a) Consider the last digits of numbers satisfying the conditions of the problem. We follow the last digit of their squares, so we put the matching the last digit to each square:

$1^2 \rightarrow 1, 2^2 \rightarrow 4, 3^2 \rightarrow 9, 4^2 \rightarrow 6, 5^2 \rightarrow 5, 6^2 \rightarrow 6, 7^2 \rightarrow 9, 8^2 \rightarrow 4, 9^2 \rightarrow 1$

b) Analogously to a), we find those single-digit numbers that satisfy the conditions and write down the last digit for each third power:

$1^3 \rightarrow 1, 2^3 \rightarrow 8, 3^3 \rightarrow 7, 4^3 \rightarrow 4, 5^3 \rightarrow 5, 6^3 \rightarrow 6, 7^3 \rightarrow 3, 8^3 \rightarrow 2, 9^3 \rightarrow 9$

- pairs $(3; 7)$ and $(2; 8)$ satisfy the conditions.

For example, we already found a pair of $l$-digit numbers $(m, n)$, satisfying the condition: $m$'s last digit is $2$ its third power ends with $n$, and analogously, number $n$'s last digit is $8$ and its third power ends with $m$. Let us try to find $l+1$-st digits of these numbers. In all equalities, all digits to the left of $l+1$-st one are discarded. We denote the unknown numbers by $\overline{xm}$ and $\overline{yn}$.

Then let $m^3 = 10' A + n$, $n^3 = 10' B + m$, which yields:

$$\begin{align*} \overline{xm}^3 &= (10' x + m)^3 \rightarrow 3 \cdot 10' x \cdot m^2 + m^3 = 3 \cdot 10' x \cdot 4 + 10' A + n \rightarrow \\ &\rightarrow 10' (2x + A) + n = \overline{yn} = 10' y + n, \\ \overline{yn}^3 &= (10' y + n)^3 \rightarrow 3 \cdot 10' y \cdot n^2 + n^3 = 3 \cdot 10' y \cdot 4 + 10' B + m \rightarrow \\ &\rightarrow 10' (2y + B) + m = \overline{xm} = 10' x + m. \end{align*}$$

After simplifications and reductions, get the system:

$$\begin{cases} 2x + A = y, \\ 2y + B = x. \end{cases}$$

From which, $x = 2y + B = 2(2x + A) + B = 4x + (2A + B)$, hence, $\begin{cases} 3x = -(2A + B), \\ 3y = -(2B + A). \end{cases}$

Since for arbitrary $A, B$ there exist corresponding digits $x, y$ (recall that the equality only applies to the last digits here), and $3x$ and $3y$ take all possible values.

Thus, corresponding pairs of numbers exist for any value of $l$.

Let us show by examples: the transition from a single-digit number to a two-digit number:

$$\begin{align*} m_1 = 2, m_1^3 = 8 &\rightarrow 10 \cdot 0 + 8 \Rightarrow A = 0, n_1 = 8, n_1^3 = 512 \rightarrow 10 \cdot 1 + 2 \Rightarrow B = 1. \\ 3x \rightarrow -(2A+B) = -1 &\rightarrow 9 \Rightarrow x = 3, 3x \rightarrow -(2B+A) = -2 \rightarrow 8 \Rightarrow y = 6. \end{align*}$$

We obtain the numbers:

$$m_2 = \overline{xm_1} = 32, m_2^3 = 32^3 = 32768, n_2 = \overline{yn_1} = 68, n_2^3 = 68^3 = 314432,$$

which satisfy the statement.

The transition from a two-digit number to a three-digit number:

$$\begin{align*} m_2 = 32, m_2^3 = 32768 &\rightarrow 100 \cdot 7 + 68 \Rightarrow A = 7. n_2 = 68, n_2^3 = 314432 \rightarrow 100 \cdot 4 + 32 \\ &\Rightarrow B = 4. \end{align*}$$

$$m_3 = \overline{xm_2} = 432, m_3^3 = 432^3 = 80621568, n_3 = \overline{yn_2} = 568, n_3^3 = 568^3 = 183250432,$$

which, too, satisfy the statement.

b) Let us prove that such pair exists for every value of $l$. We show that there is such positive integer $x$ that $2^l \mid x^2 - 1$ and $5^l \mid x^2 + 1$.

For the first divisibility, we set $x \equiv 1 \pmod{2^l}$. Using Method of Mathematical Induction (MMI) for $l$, we show that for any $l$ there exists $x$ such that $5^l \mid x^2 + 1$. If $l = 1$, then we set $x = 2$.

Suppose for some $l$, the statement holds. Consider $l+1$: if $x$ works for $l$, i.e. $x^2+1=t \cdot 5^l$, then we set $x_1 = x+k \cdot 5^l$; so

$$x_1^2 + 1 = x^2 + k^2 \cdot 5^{2l} + 2xk \cdot 5^l + 1 = t \cdot 5^l + k^2 \cdot 5^{2l} + 2xk \cdot 5^l = 5^l(t + k^2 \cdot 5^l + 2xk).$$

Obviously, there exists such $k$ that the expression in parentheses is divisible by $5$, we choose such $x_1$ for this $k$, and the statement is proved.

Then, by the Chinese remainder theorem, we select $x$ which satisfies both conditions. Then, we show that the last $l$ digits of the pair $x$ and $x^3$ satisfy the statement. Firstly, by the construction, both numbers have no more than $l$ digits. Secondly, they have different last digits: otherwise, $x^3 - x$ would be divisible by $5$, but $x^3 - x = x(x^2 - 1) \equiv -2x \pmod 5$ hence, $x$ must have been divisible by $5$, which, obviously, is impossible if $x^2 + 1 = t \cdot 5^l$.

It is left to show that $x - x^3 \equiv 0 \pmod{10^l}$, which is obvious by construction, and also

$$(x^3)^3 - x = x^9 - x = x(x^4 + 1)(x^2 + 1)(x^2 - 1) \equiv 0 \pmod{10^l},$$

which is obvious, as well.