(i) True

Explanation: we have, A = {3, {4, 5}, 6}

Let {4,5} = x

Now, A = {3, x, 6}

4,5 is not in A, {4,5} is an element of A and element cannot be subset of set,thus {4, 5} ⊄A.

(ii) True

Explanation: we have, A = {3, {4, 5}, 6}

Let {4,5} = x

Now, A = {3, x, 6}

Now, x is in A.

So, x ∈ A.

Thus, {4, 5} ϵ A

(iii) True

Explanation: {4,5} is an element of set {{4,5}}.

Let {4,5} = x

{{4,5}} = {x}

we have, A = {3, {4, 5}, 6}

Now, A = {3, x, 6}

So, x is in {x} and x is also in A.

So , {x} is a subset of A.

Hence, {{4, 5}} ⊆A

(iv) False

Explanation: 4 is not an element of A.

(v) True

Explanation: 3 is in {3} and also 3 is in A.

(vi) False

Explanation: ϕ is an element in { ϕ} but not in A.

Thus, {ϕ} ⊄ A

(vii) True

Explanation: ϕ is a subset of every set.

(viii) False

Explanation: we have, A = {3, {4, 5}, 6}

Let {4,5} = x

Now, A = {3, x, 6}

4,5 is in {3,4,5} but not in A, thus {3,4, 5} ⊄A.

(ix) True

Explanation: 3,6 is in {3,6} and also in A, thus {3, 6} ⊆A.