The greatest integer less than or equal to the number $R=(8+3\sqrt{7})^{20}$

Your approximation isn't a bad one, but it's not accurate enough to determine (1) and (2). Instead, note that $$R' = (8 + 3\sqrt{7})^{20} + (8 - 3\sqrt{7})^{20}$$ is an integer, and the latter term is close to $0$. Expand $R'$ above to determine $R'$, and thus $[R]$, mod $2$. Parts $3$ and $4$ follow from the same sort of approach.

Hints For (1), (2): Using the binomial expansion twice gives that \begin{align}A := R + (8 - 3 \sqrt{7})^{20} &= (8 + 3 \sqrt{7})^{20} + (8 - 3 \sqrt{7})^{20} \\ &= \sum_{k = 0}^{20} {20 \choose k} 8^{20 - k} (3 \sqrt{7})^k + \sum_{k = 0}^{20} {20 \choose k} 8^{20 - k} (-3 \sqrt{7})^k \\ &= \sum_{k = 0}^{20} {20 \choose k} 8^{20 - k} (1 + (-1)^k) (3 \sqrt{7})^k .\end{align} The appearance of the factor $1 + (-1)^k$ means that summands with odd $k$ are zero, so only the even terms contribute, and we can rewrite the sum as $$A = \sum_{j = 0}^{10} {20 \choose 2j} 8^{20 - 2j} (3 \sqrt{7})^{2j} = \sum_{j = 0}^{10} {20 \choose 2j} 64^{10 - j} 63^j .$$ In particular, $A$ is an integer. On the other hand, since $49 < 63 < 64$, we have $7 < 3 \sqrt{7} < 8$ and hence $0 < 8 - 3 \sqrt{7} < 1$.

For (3): Note that $(8 + 3 \sqrt{7})(8 - 3 \sqrt{7}) = 64 - 63 = 1.$

Additional hints For (1)-(2): So, the second summand of $A$ satisfies $0 < (8 - 3 \sqrt{7})^{20} < 1$. (In fact, it is very close to zero.) So, $A - 1 < R < A$, and in particular, $\lfloor R \rfloor = A - 1$. Since we can determine the parity of $A$ from the last summation expression, we can also determine that of $\lfloor R \rfloor$. For (3): So $$(8 + 3 \sqrt{7})^{20} (8 - 3 \sqrt{7})^{20} = 1 ,$$ hence $$(8 - 3 \sqrt{7})^{20} = \frac{1}{R} .$$

There appears to be a typo in the equation in (4).

Building on Anomaly's answer, let $a_n = (8 + 3\sqrt{7})^n + (8 - 3 \sqrt{7})^n$. Prove that $a_n$ can be defined by induction by $a_0 = 2$, $a_1 = 16$, $a_n = 16a_{n-1} - a_{n-2}$. This will help in obtaining information about $a_{20}$.