Correct : a

Given Grammar:
The grammar G is:
S → F  |  H
F → p  |  c
H → d  |  c
Where S, F, H are non-terminals, and p, d, c are terminals.
Check for LL(1) Property (Statement S1):
A grammar is LL(1) if, for every non-terminal A with productions A → α₁ | α₂ | ... | α_k, the following conditions hold:
• For any two distinct productions A → α_i and A → α_j, the sets FIRST(α_i) and FIRST(α_j) must be disjoint (i.e., FIRST(α_i) ∩ FIRST(α_j) = ∅).
• If ε ∈ FIRST(α_i), then FIRST(α_j) ∩ FOLLOW(A) = ∅ for all j ≠ i.

Let's compute the FIRST sets for the productions:
• FIRST(p) = {p}
• FIRST(c) = {c}
• FIRST(d) = {d}

• For non-terminal F:
    F → p     → FIRST(p) = {p}
    F → c     → FIRST(c) = {c}
    So, FIRST(F) = {p, c}

• For non-terminal H:
    H → d     → FIRST(d) = {d}
    H → c     → FIRST(c) = {c}
    So, FIRST(H) = {d, c}

• For non-terminal S:
    S → F     → FIRST(F) = {p, c}
    S → H     → FIRST(H) = {d, c}

Now, let's check the LL(1) condition for S:
• We have productions S → F and S → H.
• We need to check if FIRST(F) ∩ FIRST(H) is empty.
• FIRST(F) = {p, c}
• FIRST(H) = {d, c}
• FIRST(F) ∩ FIRST(H) = {c}

Since the intersection {c} is not empty, the grammar fails the LL(1) condition for non-terminal S. An LL(1) parser would not know whether to choose the production S → F or S → H when the next input symbol is 'c'.
Therefore, statement S1 is incorrect.
Check for LR(1) Property (Statement S2):
A grammar is LR(1) if its LR(1) parsing table has no shift/reduce or reduce/reduce conflicts. A key characteristic that often leads to LR conflicts is ambiguity. If a grammar is ambiguous, it cannot be LR(k) for any k.
Let's consider the string "c".
• One derivation for "c" is: S ⇒ F ⇒ c
• Another derivation for "c" is: S ⇒ H ⇒ c

Since the string "c" has two distinct leftmost derivations (or parse trees), the grammar is ambiguous.

An ambiguous grammar cannot be LR(1). During LR parsing, when the parser has processed 'c' and is about to reduce, it would be in a state with items like `F → c.` (reduce by F → c) and `H → c.` (reduce by H → c). If the lookahead symbol (which would be '$' for the entire string "c") is common to the lookahead sets of both reduce items, a reduce/reduce conflict occurs. In this case, both reductions would have '$' as a lookahead, leading to a conflict.

Therefore, statement S2 is incorrect.
Final Conclusion:
Since both S1 and S2 are incorrect, the correct option is Neither S1 and S2.