How to solve \$(X+Y)(\bar X+\bar Z)\$?
There is nothing to "solve". You have a formula of Boolean algebra that you have rearranged from a product of sums, to a sum of products.

To "solve" usually means to determine values for the variables. That requires some constraints. You need to turn the formula into an equation. And possibly you either need a additional equations, or to be given the values of some of the variables (so that only one remains unknown).

If we assume that the equation to be solved is this:

$$(X+Y)(\bar X+ \bar Z) = 1$$

In other words, "over what domain values is this formula true?" then the solution is a set: it consists of the set of \$X\$, \$Y\$ and \$Z\$ triplets which satisfy the equation.  We can obtain this solution set by writing down the truth table.

In an answer to this stack overflow question I made a truth-table generating program, which we can just run from our Linux prompt:


$ txr -i truth.tl
1> (pretty-truth-table '((x or y) and (not x or not z)))
    x     y     z   | (x or y) and (not x or not z)
--------------------+------------------------------
    F     F     F   |               F
    F     F     T   |               F
    F     T     F   |               T
    F     T     T   |               T
    T     F     F   |               T
    T     F     T   |               F
    T     T     F   |               T
    T     T     T   |               F
nil


From this we can read out the solutions: all the \$X\$, \$Y\$ and \$Z\$ values from the variable columns, which have a T in the formula column.

Note that this is equivalent to just converting the formula to a sum of products.

Can we find a simpler formula which computes the same function over the variables? For that, we can use the Karnaugh Map technique, since there are only three variables. Since we have the truth table, it's easy to stuff the map:


     Z  0  1
X Y  +------
     |
0 0  |  0  0
     |
0 1  |  1  1
     |
1 1  |  1  0
     |
1 0  |  1  0
     |


And here is how we can play it:


     Z  0  1
X Y  +------
     |
0 0  |  0  0
     | +----+
0 1  | |1  1|    X'Y term  (Z is irrelevant)
     | +----+
1 1  |  1  0
     | 
1 0  |  1  0
     |


     Z  0  1
X Y  +------
     |
0 0  |  0  0
     |
0 1  |  1  1
     | +-+------ XZ' term (Y is irrelevant)
1 1  | |1| 0
     | | |
1 0  | |1| 0
     | +-+


We end up with \$\bar XY + X\bar Z\$. This is not really simpler than the original, because it still has four variables, three binary operators and two unary negations. We can draw the syntax trees to prove that the complexity is exactly the same:



  original:                      new:

        and                               or
      /     \                           /     \
   or        or                     and         and
  /   \     /   \                  /   \       /    \
  X   Y    not  not               not   Y      X    not
            |    |                 |                 |
            X    Z                 X                 Z


The raw, syntactic complexity of the best expression we can find with a Karnaugh map is exactly the same as the original.

That said, there is some semantic simplification in the sum-of-products representation in that it exposes clearly the regions of the XYZ variable domain. It gives use an "at a glance" view over that space, telling us where the expression is true, which is useful.