Solved the first problem using common lisp.

1/sum-integers.lisp

@@ -0,0 +1,23 @@
+(in-package :nisp.euler)
+
+(defun sum-integers-if (predicate start end &optional (step 1))
+  "Sum numbers satisfying PREDICATE from START to END."
+  (iterate (for n :from start :to end :by step)
+           (when (funcall predicate n)
+             (sum n :into result))
+           (declare (fixnum result))    ; Help the compiler optimize.
+           (finally (return result))))
+
+(defun solution-1/1 ()
+  "Call `sum-integers-if' after we compose predicate.
+
+This is a purely functional way of solving this problem."
+  (sum-integers-if (disjoin (compose (curry #'= 0) (rcurry #'mod 3))
+                            (compose (curry #'= 0) (rcurry #'mod 5)))
+                   0 999))
+
+(defun solution-1/2 ()
+  "Use a local function definition for the predicate."
+  (flet ((validp (n)
+           (or (= 0 (mod n 3)) (= 0 (mod n 5)))))
+    (sum-integers-if #'validp 0 999)))

nisp.euler.asd

@@ -0,0 +1,7 @@
+(asdf:defsystem :nisp.euler
+  :serial t
+  :components
+  ((:file "package")
+   (:module "1"
+            :components
+            ((:file "sum-integers")))))

package.lisp

@@ -0,0 +1,3 @@
+(defpackage #:nisp.euler
+  (:use :cl :iterate :alexandria)
+  (:nicknames :euler))