@Arjun sir already provided the best answer for this but this is another way of solving this. Strategy behind these question is assume L.H.S as true and make R.H.S as false by some values for which L.H.S is true-

1)`((∀x)[α]⇒(∀x)[β])⇒(∀x)[α⇒β]`

Assume some values of x for α and β which makes LHS as true-

In LHS (∀x)[α] is false for the assumed domain and same as (∀x)[β] is false for the assumed domain. Now we know that *F-> F = T *makes LHS true.

In RHS [α⇒β] is false for x1 and true for x2. So for all x RHS will become false.

$\therefore$ **T -> F = F** (Not Valid)

2)`(∀x)[α]⇒(∃x)[α∧β]`

Assume some values of x for α and β which makes LHS as true-

In LHS (∀x)[α] is true for the assumed domain which makes the whole LHS true.

In RHS [α$\wedge$β] is false for x1 as well as for x2. So in RHS there is no true value, which makes whole RHS as false.

$\therefore$ **T -> F = F** (Not Valid)

3)`((∀x)[α∨β]⇒(∃x)[α])⇒(∀x)[α]`

Assume some values of x for α and β which makes LHS as true-

In LHS [α∨β] is true for the x1 as well as for x2 which makes (∀x)[α∨β] as true and (∃x)[α] is true for the assumed domain because of x1. Now we know that *T-> T = T *makes LHS true.

In RHS (∀x)[α] is false for the assumed domain which makes the whole RHS false.

$\therefore$ **T -> F = F** (Not Valid)

4)`(∀x)[α⇒β]⇒(((∀x)[α])⇒(∀x)[β])`

Assume some values of x for α and β which makes LHS as true-

x |
α |
β |

x1 |
T |
T |

x2 |
F |
T |

x3 |
F |
F |

In LHS [α⇒β] is true for x1,x2 and x3. So for all x LHS will become true.

In RHS (∀x)[α] is false for the assumed domain and same as (∀x)[β] is false for the assumed domain. Now we know that *F-> F = T *makes RHS true.

$\therefore$ **T-> T = T** (Valid)

So correct answer is option **D.**