4.
(i) Solution: Since
$$(258,348)=6,$$

and $6 \nmid 131$, so by Lemma 11, the congruence has no solution.
(ii) Solution: Since
$$29=9 \times 3+2, \quad 3=2+1$$

Therefore,
$$1=3-2=3-(29-9 \times 3)=10 \times 3-29$$

That is,
$$3 \times 10 \equiv 1(\bmod 29)$$

From the original equation,
$$3 \times 10 x \equiv 10 \times 10 \quad(\bmod 29)$$

Thus,
$$x \equiv 100 \equiv 13 \quad(\bmod 29)$$
(iii) Solution: Since
$$\begin{array}{l} 111=2 \times 47+17, \quad 47=2 \times 17+13 \\ 17=13+4, \quad 13=3 \times 4+1 \end{array}$$

Therefore,
$$\begin{aligned} 1= & 13-3 \times 4=13-3 \times(17-13) \\ = & 4 \times 13-3 \times 17=4 \times(47-2 \times 17)-3 \times 17 \\ = & 4 \times 47-11 \times 17=4 \times 47-11 \times(111 \\ & -2 \times 47)=26 \times 47-11 \times 111 \end{aligned}$$

That is,
$$26 \times 47 \equiv 1 \quad(\bmod 111)$$

From the original equation,
$$26 \times 47 x \equiv 26 \times 89(\bmod 111)$$

Thus,
$$x \equiv 26 \times 89 \equiv 94(\bmod 111)$$
(iv) Solution: Since
$(660,1385)=5$, and $5 \mid 595$, so by Exercise 3, the equation has five solutions. Now, we first solve
$$132 x \equiv 119(\bmod 277)$$

Since
$$\begin{array}{l} 277=2 \times 132+13, \quad 132=10 \times 13+2 \\ 13=6 \times 2+1 \end{array}$$

Therefore,
$$\begin{aligned} 1 & =13-6 \times 2=13-6 \times(132-10 \times 13) \\ & =61 \times 13-6 \times 132=61 \times(277-2 \\ & \times 132)-6 \times 132=61 \times 277-128 \times 132 \end{aligned}$$

That is,
$$-128 \times 132 \equiv 1(\bmod 277)$$

From the original equation,
$$-128 \times 132 x \equiv-128 \times 119(\bmod 277)_{2}$$

Thus,
$$x \equiv-128 \times 119 \equiv 3 \quad(\bmod 277)$$

The five solutions to the original congruence are
$$\begin{aligned} x & \equiv 3(\bmod 1385) \\ x & \equiv 280(\bmod 1385) \\ x & \equiv 557(\bmod 1385) \\ x & \equiv 834 \quad(\bmod 1385) \\ x & \equiv 1111 \quad(\bmod 1385) \end{aligned}$$