HINT: $L_1\cup L_2=L_2$, and $L_1\cap L_2=L_1$. For (d), note that you can design a DFA that handles the members of any finite collection of words, like $\{\lambda,ab,a^2b^2\}$, individually, and handles all other words separately.
Your use of the pumping lemma in (a) is correct.
The pumping lemma for regular languages states: If $L$ is regular then there exists an $n > 0$, such that for any $x \in L$ with $|x| \geq n$ there exist strings $u, v, w$ such that $x = uvw$, $|uv| \leq n$, $|v| > 0$ and for all $m \geq 0$, $uv^mw \in L$.
The pumping lemma for context free languages states: If $L$ is context free then there exists an $n > 0$ such that for any $x \in L$ with $|x| \geq n$ there exist strings $v, w, x, y, z$ such that $u = vwxyz$, $|wxy| \leq n$, $|wy| > 0$ and for all $m\geq 0$, $uw^mxy^mz \in L$.
Now indeed, if a language $L$ satisfies the conclusion of the pumping lemma for regular languages, then it also satisfies the conclusion of the pumping lemma for context free languages by picking $v = w = \epsilon$ and $x = u$ and $y = v$ and $z = w$, where $u, v, w$ are strings from the conclusion of the pumping lemma for regular languages.
And so what you say is indeed correct. If the conclusion of the pumping lemma for context free languages fails for a particular language $L$ then the language cannot be regular, because the conclusion of the pumping lemma for regular languages must fail for $L$.
However this also follows from the fact that any regular language is context free and in fact can be given by a very simple grammar, called a regular grammar. So if the conclusion of the pumping lemma for context free languages fails, then the language is not context free, which also means that it is not regular.
Best Answer
Say that $m$ satisfies the pumping property for a language $L$ if the following holds :
Let $m := \max(n_1, n_2)$. We show that $m$ satisfies the pumping property for $L_1 \cup L_2$ :
Indeed, let $w \in L_1 \cup L_2$, then either $w \in L_1$ or $w \in L_2$. If $w \in L_1$, then since $n_1$ satisfy the pumping property for $L_1$, $m$ also satisfies the pumping property for $L_1$ (since $n_1 \leqslant m$). Hence we can write $w=xyz$ as required. Likewise if $w \in L_2$. QED
Now, $n$ just happens to be the minimum of all $p$s satisfying the pumping property for $L_1 \cup L_2$, hence $n \leqslant m = \max(n_1, n_2)$.