--- 1 ---
|= F<->G iff
|= F->G and |= G->F iff (by deduction theorem)
F |= G and G |= F iff
Models(F) \subseteq Models(G) and Models(G) \subseteq Models(F) iff
Models(F) = Models(G) iff
F equiv G

--- 2 ---
Express all connectives by - and &.
E.g.:
A | B = -(-A & -B) by De Morgan's laws
A -> B = -A | B = -(A & -B)
...

Other minimal functionally complete sets:
{or, not}
{imp, not}

nand is functionally complete
so is nor

--- 3 ---
(a)
a&b | c&d =
(a&b | c) & (a&b | d) =
(a|c) & (b|c) & (a|d) & (b|d)

(c)
d g -sp -st |
d g -sp r |
d g g -st |
d g g r |
d -ra -sp -st |
d -ra -sp r |
d -ra g -st |
d -ra g r

proof:
d & (g|-ra) & (-sp|g) & (-st|r) =
(d&g) | (d & -ra) & (-sp|g) & (-st|r) =
(d&g) | (d & -ra) & (-sp|g) & (-st|r) =
(d&g&(-sp|g) | d&-ra&(-sp|g))& (-st|r) =
(d&g&-sp| d&g&g | d&-ra&-sp| d&-ra&g)& (-st|r) =
(d&g&-sp& (-st|r) | d&g&g& (-st|r) | d&-ra&-sp& (-st|r) | d&-ra&g& (-st|r)) =
starting formula

--- 4 ---
d <-
g <- ra
g <- sp
r <- st

M0 = {}
M1 = T(M0) = {d}
M2 = T(M1) = {d} = M

Then g does not hold.

--- 5 ---
(. stands for open box)
a)

1 . P & -P (assumption)
2 . P (and elimination 1)
3 . -P (and elimination 1)
4 . bottom (contradiction 2+3)
5 -(P & -P) (negation introduction 1+4)

b)

1 -P
2 . -Q (assumption)
3 .. P (assumption)
4 .. bottom (contradiction 1+3)
5 . -P (negation introduction 3+4)
6 P->Q (contraposition 2+5)