In both cases you’re starting in the wrong place, translating the original statement into symbols incorrectly.
For (d) the original statement is essentially ‘There does not exist a dog that can talk’, i.e., $\neg\exists xP(x)$, where $P(x)$ is ‘$x$ is a dog that can talk’. Negating that gives you simply $\exists xP(x)$, ‘There is a dog that can talk’.
Similarly, assuming that the universe of discourse is this class, (e) is $\neg\exists\big(F(x)\land R(x)\big)$, where $F(x)$ is ‘$x$ does know French’ and $R(x)$ is ‘$x$ does know Russian’, so its negation is $\exists\big(F(x)\land R(x)\big)$ ‘There is someone in this class who knows French and Russian’.
here are some basic distributive properties in quantifiers, hope it might help someone.
∀x(P(x) ∧ Q(x)) ≡ (∀xP(x) ∧ ∀xQ(x))
∃x(P(x) ∧ Q(x)) → (∃xP(x) ∧ ∃xQ(x))
∀x(P(x) ∨ Q(x)) ← (∀xP(x) ∨ ∀xQ(x))
∃x(P(x) ∨ Q(x)) ≡ (∃xP(x) ∨ ∃xQ(x))
∀x(P(x) → Q(x)) ← (∃xP(x) → ∀xQ(x))
∃x(P(x) → Q(x)) ≡ (∀xP(x) → ∃xQ(x))
∀x¬P(x) ≡ ¬∃xP(x)
∃x¬P(x) ≡ ¬∀xP(x)
∀x∃yT(x,y) ← ∃y∀xT(x,y)
∀x∀yT(x,y) ≡ ∀y∀xT(x,y)
∃x∃yT(x,y) ≡ ∃y∃xT(x,y)
∀x(P(x) ∨ R) ≡ (∀xP(x) ∨ R)
∃x(P(x) ∧ R) ≡ (∃xP(x) ∧ R)
∀x(P(x) → R) ≡ (∃xP(x) → R)
∃x(P(x) → R) → (∀xP(x) → R)
∀x(R → Q(x)) ≡ (R → ∀xQ(x))
∃x(R → Q(x)) → (R → ∃xQ(x))
∀xR ← R
∃xR → R
The following formulas are not valid.
A B counterexample
∃x(P(x) ∧ Q(x)) ← (∃xP(x) ∧ ∃xQ(x)) D = {a, b}, M = {P(a), Q(b)}
∀x(P(x) ∨ Q(x)) → (∀xP(x) ∨ ∀xQ(x)) D = {a, b}, M = {P(a), Q(b)}
∀x(P(x) → Q(x)) → (∃xP(x) → ∀xQ(x)) D = {a, b}, M = {P(a), Q(a)}
∀x∃yT(x,y) → ∃y∀xT(x,y) D = {a, b}, M = {T(a,b), T(b,a)}
∃x(P(x) → R) ← (∀xP(x) → R) D = Ø, M = {R}
∃x(R → Q(x)) ← (R → ∃xQ(x)) D = Ø, M = Ø
∀xR → R D = Ø, M = Ø
∃xR ← R D = Ø, M = {R}
Note: if empty domains are not allowed, then the last four implications are in fact valid.
Best Answer
The variables stand for properties (or propositions). The predicate $T$ is for tautologies, i.e., $T(x)$ means that the property $x$ is a tautology. $C(x)$ means that $x$ is a contradiction.
Now things should be rather straight forward:
$\exists x T(x)$ means "there is a property $x$ such that $x$ is a tautology".
The second line is "for all propositions $x$ such that $x$ is a contradiction, the negation $\neg x$ is a tautology".
The third line is more interesting. I believe the last $\wedge$ should be $\vee$ for disjunction. Then the line can be explained as follows:
What is a contingency? A property that is neither a tautology nor a contradiction. So this line says "there are $x$ and $y$ such that $x$ and $y$ are contingencies and $x\vee y$ is a tautology".
The last line also has a typo, I think. The comma between $x$ and $y$ should be $\wedge$.