First approach. Suppose that the LHS of (i) is false for some triple of positive numbers $a$, $b$, $c$ that satisfy (1), so that $1 = 2abc + ab + bc + ca$, but $ab + bc + ca < 3/4$. Then, by the AM-GM inequality,
$$\begin{aligned} abc &= \sqrt{(ab)(bc)(ca)} = (\sqrt[3]{(ab)(bc)(ca)})^{3/2} \\ &\leq \left( \frac{ab + bc + ca}{3} \right)^{3/2} < \left( \frac{1}{4} \right)^{3/2} = \frac{1}{8}, \end{aligned}$$
whence $2abc + ab + bc + ca < \frac{2}{8} + \frac{3}{4} = 1,$
in contradiction to our assumption. Hence, (i) holds, and (ii) now follows as an immediate consequence, because from (i) and (1) we see that
$$2xyz = 1 - (xy + yz + zx) \le 1 - \frac{3}{4} = \frac{1}{4} \iff xyz \le \frac{1}{8}.$$

Second approach. The proof of the LHS of (i) can be done directly as follows. By the AM-GM inequality,
$$xyz = \left( \sqrt[3]{(xy)(yz)(zx)} \right)^{3/2} \le \left( \frac{xy + yz + zx}{3} \right)^{3/2},$$
and so, with $t = (xy + yz + zx)/3$, we have that
$$1 = 2xyz + xy + yz + zx \le 2t^{3/2} + 3t \iff 1 \le 2(\sqrt{t})^3 + 3(\sqrt{t})^2,$$
i.e., letting $s = \sqrt{t}$, $r = 2s$
$$1 \le 2s^3 + 3s^2 \iff 4 \le r^3 + 3r^2 \iff 0 \le (r-1)(r+2)^2 \iff r \ge 1.$$
Thus, $t \ge 1/4$, i.e., $xy + yz + zx \ge 3/4$, as claimed.

Here is a direct way to prove (ii). By the AM-GM inequality,
$$xy + yz + zx \ge 3\sqrt[3]{(xy)(yz)(zx)} = 3(xyz)^{2/3} = 3s^2 \quad (\text{where } s = \sqrt[3]{xyz}).$$
Hence $1 = 2xyz + xy + yz + zx \ge 2s^3 + 3s^2$ and this is equivalent to
$$0 \ge 2(s^3 + 1) + 3(s^2 - 1) = (s+1)^2(2s-1) \iff 2s \le 1.$$
Equivalently, $xyz = s^3 \le 1/8$. Again, (i) follows from this.

To prove (2), note that
$$(x + y + z)^2 \geq 3(xy + yz + zx) \geq \frac{9}{4} \iff x + y + z \geq \frac{3}{2},$$
with equality iff
$$\frac{9}{4} = (x + y + z)^2 = 3(xy + yz + zx),$$
i.e., $x + y + z = 3/2$, and $x = y = z$, i.e., $x = y = z = 1/2$.