= Solution
Write $D_k=M_k-M_{k-1}$ for the <martingale difference sequence>. <Bounded increments> mean that one deterministic $C<\infty$ satisfies $|D_k|\leq C$ for every $k$, almost surely. Then $\mathbb E D_k=0$ and, for $j<k$,
$$
\mathbb E[D_jD_k]
=\mathbb E\left[D_j\mathbb E[D_k\mid\mathcal F_{k-1}]\right]=0.
$$
The increments are <uncorrelated random variables>, uniformly bounded in $L^2$. Applying the supplied <strong law for uniformly L2-bounded uncorrelated random variables> gives $n^{-1}\sum_{k=1}^nD_k\to0$ almost surely. Also $M_0/n\to0$, because $M_0$ is finite almost surely. Therefore \b[a martingale with <bounded increments> has zero linear growth]:
$$
\boxed{\frac{M_n}{n}\longrightarrow0\quad\text{almost surely}.}
$$
This is the <strong law for martingales with bounded increments>.
For the <submartingale>, let $\Delta Y_k=Y_k-Y_{k-1}$ and define its predictable drift and compensator by
$$
a_k=\mathbb E[\Delta Y_k\mid\mathcal F_{k-1}]\geq0,
\qquad A_n=\sum_{k=1}^na_k,
\qquad N_n=Y_n-A_n.
$$
This is the <Doob decomposition in discrete time>. The process $N_n$ is a <martingale>. Since $|\Delta Y_k|\leq C$, we have $0\leq a_k\leq C$, so $|N_k-N_{k-1}|\leq2C$. The result just proved gives $N_n/n\to0$ almost surely, and $A_n\geq0$. Consequently the <strong law for submartingales with bounded increments> is
$$
\boxed{\liminf_{n\to\infty}\frac{Y_n}{n}\geq0\quad\text{almost surely}.}
$$
The compensator may grow, so the conclusion is a lower bound rather than a claim of convergence to zero.
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