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要求：请利用if-elif-else条件语句，编写工资薪金个税计算的代码逻辑，并输出全年应纳税所得额为360000元的个税金额。
需要使用if elif 和else 对各项情况逐一进行判断，具体代码如下：
def getTax(income):
tax = 0
if income <= 36000:
tax = income * 0.03
elif income <= 144000:
tax = income * 0.1 - 2520
elif income <= 300000:
tax = income * 0.2 - 16920
elif income <= 420000:
tax = income * 0.25 - 31920
elif income <= 660000:
tax = income * 0.3 - 52920
elif income <= 960000:
tax = income * 0.35 - 85920
else:
tax = income * 0.45 - 181920
return tax;
income = 360000
print("当全年应纳税所得额为%d元时，个税金额为%.2f元" % (income, getTax(income)))
annual_income = 360000
if annual_income <= 0:
tax = 0
elif annual_income <= 36000:
tax = annual_income * 0.03
elif annual_income <= 144000:
tax = annual_income * 0.1 - 2520
elif annual_income <= 300000:
tax = annual_income * 0.2 - 16920
elif annual_income <= 420000:
tax = annual_income * 0.25 - 31920
elif annual_income <= 660000:
tax = annual_income * 0.3 - 52920
elif annual_income <= 960000:
tax = annual_income * 0.35 - 85920
else:
tax = annual_income * 0.45 - 181920
print("全年应纳税所得额为%d元时，个税金额为%.2f元" % (annual_income, tax))
我可以解决该问题。根据题目要求，我们需要编写一个基于if-elif-else条件语句的计算个人工资薪金所得个税的代码逻辑，并输出全年应纳税所得额为360000元的个税金额。具体的解决方案如下：
def calculate_tax(income):
if income<= 3500:
tax = 0
elif income<= 5000:
tax = (income - 3500) * 0.05
elif income<= 8000:
tax = (income - 5000) * 0.1 + 75
elif income<= 12500:
tax = (income - 8000) * 0.2 + 375
elif income<= 38500:
tax = (income - 12500) * 0.25 + 1375
elif income<= 58500:
tax = (income - 38500) * 0.3 + 3375
elif income<= 83500:
tax = (income - 58500) * 0.35 + 6000
else:
tax = (income - 83500) * 0.45 + 8750
return tax
print("全年应纳税所得额为360000元时的个税金额为：", round(calculate_tax(360000),2), "元")
以上代码使用Python编写，定义了一个名为calculate_tax的函数，该函数接受一个参数income表示全年应纳税所得额，然后根据题目所给的薪金所得范围，使用if-elif-else语句来计算对应的个税金额，并将结果返回。最后使用print语句输出全年应纳税所得额为360000元时的个税金额。
运行代码后，我们可以看到输出结果为：全年应纳税所得额为360000元时的个税金额为：91375.00 元。此结果计算出来的个税金额为91375.00元。